UNIVERSITY  OF  CALIFORNIA 
AT   LOS  ANGELES 


A  TKXT^BOOK: 


MECHANICS  OF  MATERIALS 


BEAMS,  COLUMNS,  AND  SHAFTS. 


BY 

MANSFIELD  MERRIMAN, 

PROFESSOR   OF  CIVIL   ENGINEERING   AT   LEHIGH   UNIVERSITY. 


NEW  YORK : 
JOHN     WILEY    &     SONS, 

15   ASTOR   PLACE. 
1885. 


COPYRIGHT,  1885, 
BY  MANSFIELD  MERRIMAN. 


PREFACE. 


The  following  pages  contain  an  elementary  course  of  study  in 
the  resistance  of  materials  and  the  mechanics  of  beams,  columns 
and  shafts,  designed  for  the  use  of  classes  in  technical  schools  and 
colleges.  It  should  be  preceded  by  a  good  training  in  mathematics 
and  theoretical  mechanics,  and  be  followed  by  a  special  study  of 
the  properties  of  different  qualities  of  materials,  and  by  detailed 
exercises  in  construction  and  design. 

As  the  plan  of  the  book  is  to  deal  mainly  with  the  mechanics 
of  the  subject,  extended  tables  of  the  results  of  tests  on  different 
kinds  and  qualities  of  materials  are  not  given.  The  attempt, 
however,  has  been  made  to  state  average  values  of  the  quantities 
which  express  the  strength  and  elasticity  of  what  may  be  called 
the  six  principal  materials.  On  account  of  the  great  variation  of 
these  values  in  different  grades  of  the  same  material  the  wisdom 
of  this  attempt  may  perhaps  be  questioned,  but  the  experience  of 
the  author  in  teaching  the  subject  during  the  past  seven  years 
has  indicated  that  the  best  results  are  attained  by  forming  at  first 
a  definite  nucleus  in  the  mind  of  the  student,  around  which  may 
be  later  grouped  the  multitude  of  facts  necessary  in  his  own 
particular  department  of  study  and  work. 

As  the  aim  of  all  education  should  be  to  develop  the  powers 
of  the  mind  rather  than  impart  to  it  mere  information,  the  author 
has  endeavored  not  only  to  logically  set  forth  the  principles  and 
theory  of  the  subject,  but  to  so  arrange  the  matter  that  students 
will  be  encouraged  and  required  to  think  for  themselves.  The 


105       210973 


IV  PREFACE. 

problems  which  follow  each  article  will  be  found  very  useful  for 
this  purpose.  Without  the  solution  of  many  numerical  exercises 
it  is  indeed  scarcely  possible  to  become  well  grounded  in  the 
theory. 

In  the  chapters  on  flexure  many  problems  relating  to  I  beams 
and  other  wrought  iron  shapes  are  presented.  The  subject  of 
continuous  beams  is  not  developed  to  its  full  extent,  but  it  is 
thought  that  enough  is  given  for  an  elementary  course.  The 
resistance  of  columns  has  been  treated  with  as  much  fullness  as 
now  appears  practicable  from  a  theoretical  point  of  view. 
Considerable  attention  has  been  paid  to  combined  stresses,  and 
particularly  to  the  combination  of  torsion  and  flexure  in  shafts. 
A  new  formula  for  the  case  of  repeated  stresses  is  offered  as  a 
substitute  for  those  of  Launhardt  and  Weyrauch.  The  attempt 
has  been  made  throughout  to  render  the  examples,  exercises  and 
problems  of  a  practical  nature,  and  also  of  a  character  to  clearly 
illustrate  the  principles  of  the  theory  and  the  methods  of 
investigation. 

MANSFIELD  MERRIMAN. 

BETHLEHEM,  PA.,  June,  1885. 


CONTENTS. 


CHAPTER  I. 

ON  THE  RESISTANCE  AND  ELASTICITY  OF  MATERIALS. 

ART.     i.    AVERAGE  WEIGHTS             ....  i 

2.  STRESSES  AND  STRAINS             ....  3 

3.  EXPERIMENTAL  LAWS         .....  5 

4.  ELASTIC  LIMIT  AND  COEFFICIENT  OF  ELASTICITY    .  6 

5.  TENSION       .......  8 

6.  COMPRESSION     .            .            .            .            .            .  12 

7.  SHEAR          .......  14 

8.  FACTORS  OF  SAFETY  AND  WORKING  STRESSES         .  16 

CHAPTER  II. 

ON  PIPES,  CYLINDERS,  AND  RIVETED  JOINTS. 

ART.    9.    WATER  AND  STEAM  PIPES             ....  20 

10.  CYLINDERS  AND  SPHERES         ....  22 

11.  RIVETED  JOINTS      ......  24 

12.  MISCELLANEOUS  EXERCISES     ....  27 

CHAPTER  III. 

ON  CANTILEVERS  AND  SIMPLE  BEAMS. 

ART.  13.    DEFINITIONS            ......  29 

14.  REACTIONS  OF  THE  SUPPORTS             ...  30 

15.  EXTERNAL  FORCES  AND  INTERNAL  STRESSES     .            .  32 

1 6.  THE  VERTICAL  SHEAR             ....  34 

17.  THE  BENDING  MOMENT     .....  37 

1 8.  THEORETICAL  AND  EXPERIMENTAL  LAWS       .           .  39 

19.  THE  RESISTING  SHEAR  AND  THE  RESISTING  MOMENT  .  40 

20.  THE  TWO  FUNDAMENTAL  FORMULAS  ...  41 

21.  CENTER  OF  GRAVITY  OF  CROSS-SECTIONS           .           .  43 

22.  MOMENT  OF  INERTIA  OF  CROSS-SECTIONS     .           .  44 

23.  THE  MAXIMUM  BENDING  MOMENT          ...  45 

24.  THE  INVESTIGATION  OF  BEAMS           ...  47 


yi  CONTENTS. 

25.  SAFE  LOADS  FOR  BEAMS  .....  48 

26.  DESIGNING  OF  BEAMS  .....  49 

27.  THE  MODULUS  OF  RUPTURE          .  .  .  .51 

28.  COMPARATIVE  STRENGTH          ...  52 

29.  RECTANGULAR  BEAMS        .  .  -53 

30.  WROUGHT  IRON  I  BEAMS        .  54 

31.  WROUGHT  IRON  DECK  BEAMS      ....  56 

32.  CAST  IRON  BEAMS         ...  57 

33.  GENERAL  EQUATION  OF  THE  ELASTIC  CURVE    .           .  59 

34.  DEFLECTION  OF  CANTILEVERS             ...  61 

35.  DEFLECTION  OF  SIMPLE  BEAMS    ....  63 

36.  COMPARATIVE  DEFLECTION  AND  STIFFNESS  .            .  65 

37.  RELATION  BETWEEN  DEFLECTION  AND  STRESS  .            .  67 

38.  RECAPITULATION            ....  67 

39.  CANTILEVERS  OF  UNIFORM  STRENGTH     .                       .  68 

40.  SIMPLE  BEAMS  OF  UNIFORM  STRENGTH         .            .  71 

CHAPTER  IV. 

ON  RESTRAINED  BEAMS  AND  ON  CONTINUOUS  BEAMS. 

ART.  41.  GENERAL  PRINCIPLES  73 

42.  BEAMS  OVERHANGING  ONE  SUPPORT   ...  76 

43.  BEAMS  FIXED  AT  ONE  END  AND  SUPPORTED  AT  THE 

OTHER           ......  78 

44.  BEAMS  OVERHANGING  BOTH  SUPPORTS           .            .  81 

45.  BEAMS  FIXED  AT  BOTH  ENDS      .  .  .  .81 

46.  COMPARISON  OF  RESTRAINED  AND  SIMPLE  BEAMS  .  83 

47.  PROPERTIES  OF  CONTINUOUS  BEAMS         ...  84 

48.  THE  THEOREM  OF  THREE  MOMENTS             .            .  87 

49.  CONTINUOUS  BEAMS  WITH  EQUAL  SPANS             .            .  88 

50.  CONTINUOUS  BEAMS  WITH  UNEQUAL  SPANS  .            .  91 

51.  REMARKS  ON  THE  THEORY  OF  FLEXURE           .           .  92 

CHAPTER  V. 

ON  THE  COMPRESSION  OF  COLUMNS. 

ART.  52.  CROSS-SECTIONS  OF  COLUMNS              ...  96 

53.  GENERAL  PRINCIPLES         ...  .98 

54.  EULER'S  FORMULAS      .                                               .  99 

55.  HODGKINSON'S  FORMULAS  .....        102 


CONTENTS.  Vll 

56.  TREDGOLD'S  FORMULA             .           .  .           .             103 

57.  GORDON'S  FORMULA  .....        106 

58.  RADIUS  OF  GYRATION  OF  CROSS-SECTIONS  .            .              109 

59.  INVESTIGATION  OF  COLUMNS          .            .  .            .no 

60.  SAFE  LOADS  FOR  COLUMNS      .            .  .            .              in 

61.  DESIGNING  OF  COLUMNS    .            .           .  .            .        in 

62.  EXPERIMENTAL  RESULTS           .            .  .            .              113 

63.  REMARKS  ON  THE  THEORY  OF  COLUMNS  .           .        113 

CHAPTER  VI. 

ON  TORSION  AND  SHAFTS  FOR  TRANSMITTING  POWER. 

ART.  64.    THE  PHENOMENA  OF  TORSION             .  .            .              118 

65.  THE  FUNDAMENTAL  FORMULA  FOR  TORSION  .            .        119 

66.  POLAR  MOMENTS  OF  INERTIA  .            .  .            .              120 

67.  THE  CONSTANTS  OF  TORSION        .           .  .           .121 

68.  SHAFTS  FOR  TRANSMISSION  OF  WORK  .            .              122 

69.  ROUND  SHAFTS       .           .           .           .  .           .123 

70.  SQUARE  SHAFTS             .            .           .  .            .              124 

71.  MISCELLANEOUS  EXERCISES           .           .  .           .125 

CHAPTER  VII. 

ON  COMBINED  STRESSES. 

ART.  72.    CASES  OF  COMBINED  STRESSES           .  .            .              126 

73.  STRESSES  DUE  TO  TEMPERATURE             .  .            .127 

74.  COMBINED  TENSION  AND  FLEXURE    .  .            .              128 

75.  COMBINED  COMPRESSION  AND  FLEXURE  .  .            .        129 

76.  SHEAR  COMBINED  WITH  TENSION  OR  COMPRESSION  132 

77.  COMBINED  FLEXURE  AND  TORSION          .  .            .134 

78.  COMBINED  COMPRESSION  AND  TORSION  .            .              137 

CHAPTER  VIII. 

APPENDIX  AND  TABLES. 

ART.  79.    HORIZONTAL  SHEAR  IN  BEAMS     .            .  .           .138 

80.  MAXIMUM  INTERNAL  STRESS  IN  BEAMS  .            .              141 

81.  THE  FATIGUE  OF  METALS             .            .  .           .143 

82.  WORKING  STRENGTH  FOR  REPEATED  STRESSES        .  145 

83.  THE  RESILIENCE  OF  MATERIALS              .  .            .149 

84.  TABLES  OF  CONSTANTS           .            .  .           .               150 


In  scientiis  ediscendis  prosunt  exempla  magis  quam  prsecepta. 

NEWTON. 


A  TEXT-BOOK 

ON    THE 

MECHANICS  OF  MATERIALS 

AND    OF 

BEAMS,  COLUMNS,  AND  SHAFTS. 


CHAPTER   I. 

ON  THE  RESISTANCE  AND  ELASTICITY  OF  MATERIALS. 

ART.  i.     AVERAGE  WEIGHTS. 

The  principal  materials  used  in  engineering  constructions  are 
timber,  brick,  stone,  cast  iron,  wrought  iron  and  steel.  The 
following  table  gives  their  average  unit-weights  and  average 
specific  gravities. 


Average  Weight. 

Average 

Material. 

Pounds  per 
Cubic  Foot. 

Kilos  per 
Cubic  Meter. 

Specific 
Gravity. 

Timber 

40 

600 

0.6 

Brick- 

125 

2  000 

2.0 

Stone 

160 

2  560 

2.6 

Cast  Iron 

450 

7  200 

7.2 

Wrought  Iron 

480 

7  700 

7-7 

Steel 

49° 

7  800 

7-8 

2  RESISTANCE    AND    ELASTICITY    OF    MATERIALS.  I. 

These  weights,  being  mean  or  average  values,  should  be  care- 
fully memorized  by  the  student  as  a  basis  for  more  precise  knowl- 
edge, but  it  must  be  noted  that  they  are  subject  to  more  or  less 
variation  according  to  the  quality  of  the  material.  Brick,  for 
instance,  may  weigh  as  low  as  100,  or  as  high  as  150  pounds 
per  cubic  foot,  according  as  it  is  soft  or  hard  pressed.  Unless 
otherwise  stated  the  above  average  values  will  be  used  in  the 
examples  and  problems  of  this  book.  In  all  engineering  refer- 
ence books  are  given  tables  showing  the  unit-weights  for  different 
qualities  of  the  above  six  principal  materials,  and  also  for  copper, 
lead,  glass,  cements  and  other  materials  used  in  construction. 

For  computing  the  weights  of  bars,  beams  and  pieces  of 
uniform  cross-section,  the  following  simple  rules  will  often  be 
found  convenient. 

A  wrought  iron  bar  one  square  inch  in  section  and  one  yard 

long  weighs  ten  pounds. 

Steel  is  about  two  per  cent  heavier  than  wrought  iron. 
Cast  iron  is  about  six  per  cent  lighter  than  wrought  iron. 
Stone  is  about  one-third  the  weight  of  wrought  iron. 
Brick  is  about  one-fourth  the  weight  of  wrought  iron. 
Timber  is  about  one-twelfth  the  weight  of  wrought  iron. 

For  example,  consider  a  bar  of  wrought  iron  1^X3  inches 
and  12  feet  long.  Its  cross-section  is  4.5  square  inches,  hence 
its  weight  is  45  X  4  =  I 80  pounds.  A  steel  bar  of  the  same 
dimensions  will  weigh  about  184  pounds,  and  a  cast  iron  bar 
about  169  pounds. 

Problem  I .  How  many  square  inches  in  the  cross-section  of  a 
wrought  iron  railroad  rail  weighing  67  pounds  per  square  yard? 
In  a  steel  rail?  In  a  wooden  beam  ? 

Prob.  2.  Find  the  weights  of  a  wooden  beam  3  X  4^/2  inches  in 
section  and  1 3  feet  long,  and  of  a  steel  bar  one  inch  in  diameter 
and  13  feet  long. 


ART.  2.  STRESSES    AND    STRAINS.  3 

ART.  2.     STRESSES  AND  STRAINS. 

A  '  stress '  is  a  force  which  acts  upon  a  body  and  tends  to 
change  its  shape.  If  a  weight  of  400  pounds  be  suspended  by  a 
rope,  the  stress  on  the  rope  is  400  pounds.  This  stress  produces 
an  elongation  of  the  rope  which  increases  until  the  internal  mo- 
lecular forces  or  resistances  are  in  equilibrium  with  the  exterior 
stress.  Stresses  are  measured  in  pounds,  tons  or  kilograms.  A 
'unit-stress'  is  the  amount  of  stress  on  a  unit  of  area;  this  is 
expressed  either  in  pounds  per  square  inch,  or  in  kilograms  per 
square  centimeter.  Thus,  if  a  rope  of  two  square  inches  cross- 
section  sustains  a  stress  of  400  pounds,  the  unit-stress  is  200 
pounds  per  square  inch,  for  the  total  stress  must  be  regarded  as 
distributed  over  the  two  square  inches  of  cross-section. 

A  '  strain '  is  the  amount  of  change  of  shape  of  a  body  caused 
by  an  applied  stress.  For  instance,  if  a  load  be  put  on  a  pillar 
its  length  is  shortened  and  the  amount  of  shortening  is  a  strain. 
So  in  case  of  the  rope,  the  amount  of  elongation  is  a  strain. 
Strains  are  generally  measured  in  inches  or  centimeters.  In  popu- 
lar language  the  word  strain  is  usually  synonymous  with  stress 
and  indicates  force,  and  is  often  so  used  in  technical  literature, 
but  in  the  strict  language  of  science  it  means  the  effect  of  the 
force  in  deforming  the  body.  The  measure  of  a  stress  is  a  weight, 
while  that  of  a  strain  is  the  length  of  a  line. 

Three  kinds  of  simple  stress  are  produced  by  forces  which 
tend  to  change  the  shape  of  a  body.  They  are, 

Tensile,  tending  to  pull  apart,  as  in  a  rope. 
Compressive,  tending  to  push  together,  as  in  a  column. 
Shearing,  tending  to  cut  across,  as  in  punching  a  plate. 

The  nouns  corresponding  to  these  three  adjectives  are  Tension, 
Compression,  and  Shear.  The  stresses  which  occur  in  beams, 
columns,  and  shafts  are  of  a  complex  character,  but  they  may 
always  be  resolved  into  the  three  kinds  of  simple  stress.  The 


4  RESISTANCE   AND    ELASTICITY    OF    MATERIALS.  I. 

first  effect  of  a  stress  is  to  cause  a  deformation,  or  strain,  in  the 
body.  This  strain  receives  a  special  name  according  to  the  kind 
of  stress  which  produces  it.  Thus, 

Tension  produces  a  tensile  strain,  or  elongation. 
Compression  produces  a  compressive  strain,  or  shortening. 
Shear  produces  a  shearing  strain,  or  detrusion. 

This  change  of  shape  is  resisted  by  the  forces  between  the 
molecules  of  the  body,  and  as  soon  as  this  internal  resistance 
balances  the  exterior  stresses  the  change  of  shape  ceases  and  the 
body  is  in  equilibrium.  But  if  the  stresses  be  increased  far 
enough  the  molecular  resistances  are  finally  overcome  and  the 
body  breaks  or  ruptures. 

Tension  and  Compression  are  similar  in  character  but  differ  in 
regard  to  direction.  A  tensile  stress  upon  a  bar  occurs  when  two 
forces  of  equal  intensity  act  upon  its  ends,  each  in  a  direction 
away  from  the  other.  In  compression  the  direction  of  the  forces 
is  reversed  and  each  acts  toward  the  bar.  Evidently  a  simple 
tensile  or  compressive  stress  upon  a  bar  is  to  be  regarded  as  evenly 
distributed  over  the  area  of  its  cross-section,  so  that  if  P  be 

the  total  stress  in  pounds  and  A  the  area  of  the  cross-section 

p 
in  inches,  the  unit-stress  is  —  in  pounds  per  square  inch. 

A 

Shear  requires  the  action  of  two  forces  exerted  in  parallel 
planes  and  very  near  together,  like  the  forces  in  a  pair  of  shears, 
from  which  analogy  the  name  is  derived.  Here  also  the  total 

shearing  stress  P  is  to  be  regarded  as  distributed  uniformly  over 

p 
the  area  A,  so  that  the  unit-stress  is  — .     And  conversely  if  5 

A 

represent  the  uniform  unit-stress  the  total  stress  P  is  A  times  £. 

In  any  case  of  simple  stress  acting  on  a  bar  let  P  be  the  total 
stress,  A  the  area  over  which  it  is  uniformly  distributed,  and  5" 
the  unit-stress.  Then, 

(i)  P=AS. 


ART.  3.  EXPERIMENTAL   LAWS.  5 

Also  let  /  be  the  total  strain  or  deformation  produced  by  the 
stress,  /  the  length  of  the  bar,  and  s  the  average  strain  per  unit 
of  length.  Then  also  may  be  written, 

(i)'  /  =  Is. 

The  laws  implied  in  the  statement  of  these  two  formulas  are 
confirmed  by  experiment,  if  the  stress  be  not  too  great. 

Unit-stress  in  general  will  be  denoted  by  .S",  whether  it  be  ten- 
sion, compression,  or  shear.  St  will  denote  tensile  unit-stress, 
Sc  compressive  unit-stress,  and  ^  shearing  unit-stress,  when  it 
is  necessary  to  distinguish  between  them. 

Prob.  3.  A  wrought  iron  rod  I  }/2  inches  in  diameter  breaks 
under  a  tension  of  67  500  pounds.  Find  the  breaking  unit-stress. 

Prob.  4.  If  a  wooden  bar  1^X3  inches  breaks  under  a  tensile 
stress  of  30  ooo  pounds,  what  stress  will  break  a  bar  2X3^ 
inches  ? 

ART.  3.     EXPERIMENTAL  LAWS. 

Numerous  tests  or  experiments  have  been  made  to  ascertain 
the  strength  of  materials  and  the  laws  that  govern  stresses  and 
strains.  The  resistance  of  a  rope,  for  instance,  may  be  investi- 
gated by  suspending  it  from  one  end  and  applying  weights  to  the 
other.  As  the  weights  are  added  the  rope  will  be  seen  to  stretch 
or  elongate,  and  the  amount  of  this  strain  may  be  measured. 
When  the  load  is  made  great  enough  the  rope  will  break,  and 
thus  its  ultimate  tensile  stress  is  known.  For  stone,  iron,  or  steel, 
special  machines,  known  as  testing  machines,  have  been  con- 
structed by  which  the  effect  of  different  stresses  on  different 
qualities  and  forms  of  materials  may  be  accurately  measured. 

All  experiments,  and  all  experience,  agree  in  establishing  the 
five  following  laws,  which  may  be  regarded  as  the  axioms  of  the 
science  of  the  strength  of  materials. 

(A) — When  a  small  stress  is  applied  to  a  body  a  small  strain 
is  produced,  and  on  the  removal  of  the  stress  the  body 


6  RESISTANCE   AND    ELASTICITY   OF    MATERIALS.  I. 

springs  back  to  its  original  form.  For  small  stresses,  then, 

materials  may  be  regarded  as  perfectly  elastic. 
(E] — Under  small  stresses  the  strains,  or  changes  of  shape, 

are    approximately    directly  proportional    to  the   forces 

which  produce  them. 
(C) — When  the  stress  is  great  enough  a  strain  is  produced 

which  is  partly  permanent,  that  is,  the  body  does  not  spring 

back  entirely  to  its  original  form  on  removal  of  the  stress. 

The   permanent  part  of  the  strain  is  termed  a  set.     In 

such  cases  the  strain  is  not  proportional  to  the  stress. 
(D) — When   the  stress  is  greater   still   the   strain    rapidly 

increases  and  the  body  finally  ruptures. 
(£) — A  sudden  stress,  or  shock,  is  more  injurious  than  a 

steady  stress  or  than  a  stress  gradually  applied. 

The  words  small  and  great,  used  in  stating  these  laws,  have, 
as  will  be  seen  later,  very  different  values  and  limits  for  different 
kinds  of  materials  and  stresses.  Let  6"  be  any  unit-stress  and  s 
the  unit-strain  produced  by  it.  Then  according  to  the  law  (E] 

s 

the  ratio  —    is  a  constant  for  small  stresses,  but  its  value  for  cast 
s 

iron  is  about  ten  times  its  value  for  timber. 

The  'ultimate  strength'  of  a  material  under  tension,  compres- 
sion, or  shear,  is  the  greatest  unit-stress  to  which  it  can  be  sub- 
jected. This  occurs  at  or  shortly  before  rupture,  and  its  value  is 
very  different  for  different  materials. 

Prob.  5.  If  a  wrought  iron  bar  i  inch  in  diameter  and  4  feet 
long  elongates  half  an  inch  under  a  certain  small  stress  P,  how 
much  will  a  bar  i  ^  inches  in  diameter  and  5  feet  long  elongate 
under  a  stress  2  P  ? 

ART.  4.     ELASTIC  LIMIT  AND  COEFFICIENT  OF  ELASTICITY. 

The  'elastic  limit'  is  that  unit-stress  at  which  the  permanent 
set  is  first  visible  and  within  which  the  stress  is  directly  propor- 
tional to  the  strain.  For  stresses  less  than  the  elastic  limit  bodies 


ART.  4.     ELASTIC    LIMIT    AND    COEFFICIENT    OF    ELASTICITY.  7 

are  perfectly  elastic,  resuming  their  original  form  on  removal  of 
the  stress.  Beyond  the  elastic  limit  a  permanent  alteration 
of  shape  occurs,  or,  in  other  words,  the  elasticity  of  the  material 
has  been  impaired.  It  is  a  fundamental  rule  in  all  engineering 
constructions  that  the  material  must  never  be  strained  beyond 
its  elastic  limit. 

The  '  coefficient  of  elasticity  '  of  a  bar  for  tension,  compression, 
or  shearing,  is  the  ratio  of  the  unit-stress  to  the  unit-strain,  pro- 
vided the  elastic  limit  of  the  material  be  not  exceeded.  Let  5  be 
the  unit-stress,  s  the  unit-strain  and  E  the  coefficient  of  elasticity. 
Then  by  the  definition, 

(2)  £=?. 

By  law  (£>)  the  quantity  E  is  a  constant  for  each  material,  until 
5  reaches  the  elastic  limit.  Beyond  this  limit  s  increases  more 
rapidly  than  5"  and  the  ratio  is  no  longer  constant.  Equation  (2) 
is  a  fundamental  one  in  the  science  of  the  strength  of  materials. 
Since  E  varies  inversely  with  s,  the  coefficient  of  elasticity  may 
be  regarded  as  a  measure  of  the  stiffness  of  the  material.  The 
stiffer  the  material  the  less  is  the  change  in  length  under  a  given 
stress  and  the  greater  is  E.  The  values  of  E  for  materials  have 
been  determined  by  experiments  with  testing  machines  and  their 
average  values  will  be  given  in  the  following  articles.  E  is 
necessarily  expressed  in  the  same  unit  as  the  unit-stress  .S. 

Another  definition  of  the  coefficient  of  elasticity  is  that  it  is 
the  unit-stress  which  would  elongate  a  bar  to  double  its  original 
length,  provided  that  it  could  be  done  without  exceeding  the 
elastic  limit.  That  this  is  in  agreement  with  (2)  may  be  shown 

by  regarding  a  bar  of  length  /  which  elongates  the  amount  ^ 

p 
under  the  unit-stress  —     Then  (2)  becomes, 

p_P_  _^_  Pl_ 

~A^~l~"  At' 

p 

and  if  X  be  equal  to  /,  E  is  the  same  as  the  unit  stress  — . 

A 


RESISTANCE   AND    ELASTICITY   OF    MATERIALS. 


T. 


Prob.  6.  Find  the  coefficient  of  elasticity  of  a  bar  of  wrought 
iron  1 1^  inches  in  diameter  and  16  feet  long  which  elongates  ^ 
inch  under  a  tensile  stress  of  1 5  ooo  pounds. 

Prob.  7.  If  the  coefficient  of  elasticity  of  cast  iron  is  1 5  ooo  ooo 
pounds  per  square  inch,  how  much  will  a  bar  2X3  inches  and  6 
feet  long  stretch  under  a  tension  of  5  ooo  pounds? 

Ans.  0.004  inches. 

ART.  5.     TENSION. 

The  phenomena  of  tension  observed  when  a  gradually  increas- 
ing stress  is  applied  to  a  bar,  are  briefly  as  follows:  When  the 
unit-stress  .S  is  less  than  the  elastic  limit  Se,  the  unit-elongation 
s  is  small  and  proportional  to  6".  Within  this  limit  the  ratio  of  6" 
to  s  is  the  coefficient  of  elasticity  of  the  material.  After  passing 
the  elastic  limit  the  bar  rapidly  elongates  and  this  is  accompanied 
by  a  reduction  in  area  of  its  cross-section.  Finally  when  5  reaches 
the  ultimate  tensile  strength  S,,  the  bar  tears  apart.  Usually  Sf 
is  the  maximum  unit-stress  on  the  bar,  but  in  some  cases  the 
unit-stress  reaches  a  maximum  shortly  before  rupture  occurs. 

The  constants  of  tension  for  timber,  cast  iron,  wrought  iron 
and  steel  are  given  in  the  following  table.  The  values  are  average 
ones  and  are  liable  to  great  variations  for  different  grades  and 
qualities  of  materials.  Brick  and  stone  are  not  here  mentioned, 
as  they  are  rarely  or  never  used  in  tension. 


Material. 

Coefficient  of 
Elasticity,  E. 

Elastic 
Limit,  Se. 

Ultimate 
Tensile 
Str'gth,  St. 

Ultimate 
Elonga- 
tion, j. 

Lbs  per  sq.  in. 

Lbs  per 

Lbs  per 

In.  per 

sq.  in. 

sq.   in. 

linear  in. 

Timber 

i  500  ooo 

3  ooo 

IO  000 

0.015 

Cast  Iron 

15  ooo  ooo 

6  ooo 

20   000 

0.005 

Wrought  Iron 

25  ooo  ooo 

25  ooo 

55  ooo 

0.15 

Steel 

30  ooo  ooo 

40  ooo 

100   000 

0.10 

The  values  of  the  coefficients  of  elasticity,  elastic   limits  and 
breaking  or  ultimate  strengths  are  given  in  pounds  per  square 


ART.  5. 


TENSION. 


inch  of  the  original  cross-section  of  the  bar.  The  ultimate  elon- 
gations are  in  fractional  parts  of  the  original  length,  or  they  are 
the  elongations  per  linear  unit ;  these  elongations,  should  be 
regarded  only  as  very  rough  averages,  since  they  are  subject  to 
great  variations  depending  on  the  shape,  size  and  quality  of 
the  specimen. 

The  ultimate  elongation,  together  with  the  reduction  in  area 
of  the  cross-section,  furnishes  the  means  of  judging  of  the  duc- 
tility of  the  material.  The  reduction  of  area  in  cast  iron  and  in 
many  varieties  of  steel  is  scarcely  perceptible,  while  in  other 
varieties  of  steel  and  in  wrought  iron  it  may  be  as  high  as  0.4  of 
the  original  section. 

A  graphical  illustration  of  the  principal  phenomena  of  tension 

is  given  in  Fig.  i.     The  unit-stresses  are  taken  as  ordinates  and 

xooooo 


Y  tjoooo 
|  BOOOO 
S  70000 
Er  6oOOO 

5s  50000 
<&  40000 
g,  30000 

•S    20OOO 


7 


Elc  ngat 


per  unit  of  1<  ngt  t=s 


,n^         O.OO         Q-Qft          OJD         0.12  OJL4- 

Fig;  i. 

the  unit-elongations  as  abscissas.  For  each  unit-stress  the  cor- 
responding unit-elongations  as  found  by  experiment  are  laid  off 
and  curves  drawn  through  the  points  thus  determined.  For  each 
of  the  materials  the  curve  is  a  straight  line  from  the  origin  until 
the  elastic  limit  is  reached,  as  should  be  the  case  according  to 
the  law  (B].  The  tangent  of  the  angle  which  this  line  makes 
with  the  axis  of  abscissas  has  evidently  the  same  value  as  the 


10 


RESISTANCE    AND    ELASTICITY    OF    MATERIALS. 


I. 


coefficient  of  elasticity  of  the  material.  At  the  elastic  limit  a 
sudden  change  in  the  curve  is  noticed  and  the  elongation  rapidly 
increases.  The  termination  of  the  curve  indicates  the  point  of 
rupture.  These  curves  show  more  plainly  to  the  eye  than  the 
values  in  the  table  can  do  the  differences  in  the  properties  of  the 
materials.  It  will  be  seen  that  the  elastic  limit  is  not  a  well 
defined  point,  but  that  its  value  is  more  or  less  uncertain,  particu- 
larly for  cast  iron  and  timber.  It  should  be  also  clearly  under- 
stood that  particular  curves  for  special  cases  would  often  show 
great  variations  from  their  mean  forms  as  represented  in 
the  diagram. 


As  a  particular  example  a 
tensile  test  of  a  wrought  iron 
bar  ^  inches  in  diameter  and 
12  inches  long  made  at  the 
Pencoyd  Iron  Works  will  be 
considered.  In  the  first  col- 
umn of  the  following  table  are 
given  the  total  stresses  which 
were  successively  applied,  in 
the  second  the  stresses  per 
square  inch,  in  the  third  the 
total  elongations,  and  in  the 
fourth  the  elongations  or  sets 
after  removal  of  the  stress. 
The  unit-elongations  are 
found  by  dividing  those  in 
the  table  by  12  inches,  the 
length  of  the  specimen. 
Then  from  formula  (2)  the 
coefficient  of  elasticity  can  be 
computed  for  different  values 
pf  6"  and  s.  Thus  for  the  second,  third  and  sixth  cases, 


Total 

Stress, 
in 
Pounds. 

Stress 
per 
Square 
Inch. 

Elongation. 

Load  on. 

Load 
off. 

2  245 

5  ooo 

.001 

.000 

4  49° 

IO   OOO 

.004 

.000 

6  735 

15  ooo 

.005 

.000 

8  980 

20  000 

.008 

.000 

9  878 

22   OOO 

.009 

.000 

10  776 

24  ooo 

.010 

.000 

ii  674 

26  ooo 

.0105 

.000 

12    572 

28  ooo 

.Oil 

.000 

13   470 

30  ooo 

.013 

.000 

14   368 

32  ooo 

.014 

.000 

15    266 

34  ooo 

.015 

.002 

16  164 

36  ooo 

.022 

.007 

17  062 

38  ooo 

.416 

•3995 

17  960 

40  ooo 

•5445 

•523 

25  450 

50  ooo 

1.740 

1.707 

23  175 

51  600 

2.468 

Specimen  broke  with  51  600  tbs.  per 

square  inch. 

Stretch  in  12  inches  2.468  inch. 

8       "       1.812     " 

8       "    22.65  per  cent. 

Fractured  area  0.297  square  inches. 

ART.  5.  TENSION.  ii 


for  S  =  15  ooo,  s  =  -      -     and  E  =  36  ooo  ooo 


for  S  =  20  ooo,  s  =          -    and  E  =  30  ooo  ooo 
12 

for  5  =  26  ooo,  s  =  -1—  -2.  and  E  =  29  700  ooo 

The  elastic  limit  was  reached  at  about  33  ooo  pounds  per 
square  inch,  indicated  by  the  beginning  of  the  set  and  the  rapid 
increase  of  the  elongations.  The  ultimate  tensile  strength  of  the 
specimen  was  51  600  pounds  per  square  inch.  The  ultimate 

unit-elongation  was    _  __  =  0.205  inches  per  linear  inch.      It 
12 

hence  appears  that  this  bar  of  wrought  iron  was  much  higher 
than  the  average  as  regards  stiffness,  elastic  limit  and  ductility, 
and  lower  than  the  average  in  ultimate  strength. 

The  'working  strength'  of  a  material  is  that  unit-stress  to 
which  it  may  safely  be  subjected.  This  should  never  be  greater 
than  the  elastic  limit  of  the  material,  since  if  that  limit  be  exceeded 
there  is  a  permanent  set  which  impairs  the  elasticity.  "  In  order 
to  secure  an  ample  margin  of  safety  it  is  customary  to  take  the 
working  strength  at  from  one-third  to  two-thirds  the  elastic  limit 
Sc.  The  reasons  which  govern  the  selection  of  exact  values  of 
the  working  strength  will  be  set  forth  in  the  following  articles. 

To  investigate  the  security  of  a  piece  subjected  to  a  tension  P, 
it  is  necessary  first  to  divide  P  by  the  area  of  the  cross-section 
and  thus  determine  the  working  strength.  Then  a  comparison 
of  this  value  with  the  value  St  for  the  given  material  will  indicate 
whether  the  applied  stress  is  too  great  or  whether  the  piece  has 
a  margin  of  safety.  For  example,  if  a  tensile  stress  of  4  500 
pounds  be  applied  to  a  wrought  iron  bar  of  ^  inches  diameter 
the  working  unit-stress  is, 

S  =  —  =  —  _  =  10  ooo  pounds  per  square  inch,  nearly. 
A,        0.449 


12  RESISTANCE   AND    ELASTICITY   OF    MATERIALS.  I. 

As  this  is  less  than  one-half  the  elastic  limit  of  wrought  iron  the 
bar  has  a  good  margin  of  security. 

To  design  a  piece  to  carry  a  given  tension  P  it  is  necessary  to 

assume  the  kind  of  material  to  be  used  and  its  allowable  working 

p 
strength  6".     Then  —  is   the   area   of  the   cross-section  of  the 

piece,  which  may  be  made  of  such  shape  as  the  circumstances  of 
the  case  require.  For  example,  if  it  be  required  to  design  a 
wooden  bar  to  carry  a  tensile  stress  of  4  500  pounds,  the  working 
strength  may  be  assumed  at  I  ooo  pounds  per  square  inch  and 
the  required  area  is  4.5  square  inches,  so  that  the  bar  may  be 
.made  2X2^  inches  in  section. 

The  elongation  of  a  bar  within  the  elastic  limit  may  readily  be 
computed  by  the  help  of  formula  (2).  For  instance,  let  it  be 
required  to  find  the  elongation  of  a  wooden  bar  3X3  inches  and 
12  feet  long  under  a  tensile  stress  of  9  ooo  pounds.  From  the 
formulas  (2)  and  (i), 

p        S          Pi  PI 

ST       A+l  ~'-AE 

Substituting  in  this  the  values  £=  i  500  ooo,  A  =  9,  1=  144 
and  P  —  9  ooo,  the  probable  value  of  the  elongation  /  is  found 
to  be  0.096  inches. 

Prob.  8.  Investigate  the  security  of  a  cast  iron  bar  2X2  inches 
when  subject  to  a  tension  of  40  ooo  pounds. 

Prob.  9.  Find  the  size  of  a  round  wrought  iron  rod  to  safely 
carry  a  tensile  stress  of  100  ooo  pounds. 

Prob.  10.  Compute  the  elongation  of  wooden  and  of  a  cast 
iron  bar,  each  being  2X3  inches  and  16  feet  long,  under  a  tensile 
stress  of  6  ooo  pounds. 

ART.  6.     COMPRESSION. 

The  phenomena  of  compression  are  similar  to  those  of  ten- 
sion, provided  that  the  length  of  the  specimen  does  not  exceed 


ART.  6. 


COMPRESSION. 


about  five  times  its  least  diameter.  The  piece  at  first  shortens 
proportionally  to  the  applied  stress,  but  after  the  elastic  limit  is 
passed  the  shortening  increases  more  rapidly  and  is  accompanied 
by  a  slight  enlargement  of  the  cross-section.  When  the  stress 
reaches  the  ultimate  strength  of  the  material  the  specimen  cracks 
and  ruptures.  If  the  length  of  the  piece  exceeds  about  ten  times 
its  least  diameter  a  sidewise  bending  or  flexure  of  the  specimen 
occurs,  so  that  it  fails  under  different  circumstances  than  those  of 
direct  compression.  All  the  values  given  in  this  article  refer  to 
specimens  whose  lengths  do  not  exceed  about  five  times  their 
least  diameter.  Longer  pieces  will  be  discussed  in  chapter  V 
under  the  head  of  'columns.'  Owing  to  the  difficulty  of  making 
experiments  on  short  specimens  and  to  an  increase  of  resistance 
that  arises  with  the  enlargement  of  the  cross-section,  the  phe- 
nomena of  compression  are  not  usually  so  regular  as  those 
of  tension. 

The  constants  of  compression  for  short  specimens  are  given  in 
the  following  table,  the  values,  like  those  for  tension,  being  rough 
average  values  liable  to  much  variation  in  particular  cases. 


Material. 

Coefficient  of 
Elasticity,  E. 

Elastic 
Limit,  Se. 

Ultimate 
Compressive 
Strength,  St. 

Lbs  per  sq.  in. 

Lbs  per  sq.  in. 

Lbs  per  sq.  in. 

Timber 

i  500  ooo 

3  ooo 

8000 

Brick 

2  500 

Stone 

6  OOO  OOO 

6000 

Cast  Iron 
Wrought  Iron 

15  ooo  ooo 

25  ooo  ooo 

25  ooo 

90000 
55000 

Steel 

30  ooo  ooo 

40000 

150000 

The  values  of  the  coefficient  of  elasticity  and  the  elastic  limit  for 
timber,  wrought  iron  and  steel  here  stated  are  the  same  as  those  for 
tension,  but  the  same  reliance  cannot  be  placed  upon  them,  owing 
to  the  irregularity  of  experiments  thus  far  made.  There  is  reason 


14  RESISTANCE   AND    ELASTICITY    OF    MATERIALS.  I. 

to  believe  that  both  the  elastic  limit  and  the  coefficient  of  elastic- 
ity for  compression  are  somewhat  greater  than  for  tension. 

The  investigation  of  a  piece  subjected  to  compression,  or  the 
design  of  a  short  piece  to  be  subjected  to  compression,  is  effected 
by  exactly  the  same  methods  as  for  tension.  Indeed  it  is  custom- 
ary to  employ  these  methods  for  cases  where  the  length  of  the 
piece  is  as  great  as  ten  times  its  least  diameter. 

Prob.  1 1 .  Find  the  height  of  a  brick  tower  which  crushes 
under  its  own  weight.  Also  the  height  of  a  stone  tower. 

Prob.  12.  The  piston  of  a  steam  engine  is  18"  in  diameter  and 
the  piston  rod  is  2".  Find  the  compressive  unit-stress  on  the 
piston  rod  when  the  steam  pressure  behind  the  piston  is  80  pounds 
per  square  inch. 

Prob.  13.  Compute  the  amount  of  shortening  in  a  wrought 
iron  specimen  i  inch  in  diameter  and  5  inches  long  under  a  load 
of  6  ooo  pounds. 

ART.  7.     SHEAR. 

Shearing  stresses  and  strains  occur  whenever  two  forces,  act- 
ing like  a  pair  of  shears,  tend  to  cut  a  body  between  them.  When 
a  plate  is  punched  the  ultimate  shearing  strength  of  the  material 
must  be  overcome  over  the  surface  punched.  When  a  bolt  is  in 
tension  the  applied  stress  tends  to  shear  off  the  head  and  also  to 
strip  or  shear  the  threads  in  the  nut  and  screw.  When  a  rivet 
connects  two  plates  which  transmit  tension  the  plates  tend  to 
shear  the  rivet  across. 

The  ultimate  shearing  strength  of  materials  is  easily  determined 
by  causing  rupture  under  a  stress  P,  and  then  dividing  P  by  the 
area  A  of  the  shorn  surface.  The  value  of  this  for  timber 
is  found  to  be  very  much  smaller  along  the  grain  than  across  the 
grain ;  for  the  first  direction  it  is  sometimes  called  longitudinal 
shearing  strength  and  for  the  second  transverse  shearing  strength. 
The  same  distinction  is  sometimes  made  in  rolled  wrought  iron 


ART.  7. 


SHEAR. 


plates  and  bars  where  the  process  of  manufacture  induces  a  more 
or  less  fibrous  structure.  The  elastic  limit  and  the  amount  of  detru- 
sion  for  shearing  are  difficult  to  determine  experimentally.  The 
coefficient  of  elasticity  however  has  been  deduced  by  means  of 
certain  calculations  and  experiments  on  the  twisting  of  shafts, 
explained  in  chapter  VI  under  the  head  of  torsion. 


Material. 

Coefficient  of 
Elasticity,  E. 

Ultimate 
Sheafing 
Strength,  Ss. 

Timber,  Longitudinal 
Timber,  Transverse 
Cast  Iron 
Wrought  Iron 
Steel 

400  ooo 

6  OOO  OOO 

15  ooo  ooo 

60O 

3  ooo 

20  000 

50  ooo 
70  ooo 

rig-.  2. 


The  investigation  and  design  of  a  piece  to  withstand  shearing 
stress  is  made  by  the  means  of  the  equation  P=  AS,  in  the  same 
manner  as  for  tension  and  compression.  For  instance,  consider  the 
cylindrical  wooden  speci- 
men shown  in  Fig.  2,  which 
has  the  following  dimen- 
sions:  length  ab  =  6 inches, 
diameter  of  ends  =  4  inches,  diameter  of  central  part  =  2  inches. 
Let  this  specimen  be  subjected  to  a  tensile  stress  in  the  direction 
of  its  length.  This  not  only  tends  to  tear  it  apart  by  tension, 
but  also  to  shear  off  the  ends  on  a  surface  whose  length  is  ab 
and  whose  diameter  is  that  of  the  central  cylinder.  The  force  P 
required  to  cause  this  longitudinal  shearing  is, 

P  =  ASS  =  3.14  X  2  X  6  X  600  =  22  600  pounds, 
while  the  force  required  to  rupture  the  specimen  by  tension  is, 

P=ASf  =  3.14  X  i2  X  10  000  =  31  400  pounds. 
As  the  former  resistance  is  only  about  two-thirds  that  of  the  latter 
the  specimen  will  evidently  fail  by  the  shearing  off  of  the  ends. 


1 6  RESISTANCE    AND    ELASTICITY    OF    MATERIALS.  I. 

Prob.  14.  A  hole  ^  inches  in  diameter  is  punched  in  a  wrought 
iron  plate  5/6  inches  thick  by  a  pressure  on  the  punch  of  78  coo 
pounds.  What  is  the  ultimate  shearing  strength  of  the  iron  ? 

Prob.  15.  A  wrought  iron  bolt  i  ^  inches  in  diameter  has  a 
head  ^  inches  long.  Find  the  unit-stress  tending  to  shear  off 
the  head  when  a  tension  of  3  coo  pounds  is  applied  to  the  bolt  ? 

ART.  8.     FACTORS  OF  SAFETY  AND  WORKING  STRESSES. 

The  factor  of  safety  for  a  body  under  stress  is  the  ratio  of  its 
ultimate  strength  to  the  actual  existing  unit-stress.  The  factor 
of  safety  for  a  piece  to  be  designed  is  the  ratio  of  the  ultimate 
strength  to  the  proper  allowable  working  strength.  Thus  if  Sf 
be  the  ultimate  and  S  the  working  strength,  the  factor  of  safety 

f  is  /= — *-.     The  factor  of  safety  is  then  always   an   abstract 

number,  which  indicates  the  number  of  times  the  working  stress 
may  be  multiplied  before  the  rupture  of  the  body. 

The  law  (£)  in  Art.  3  indicates  that  working  stresses  should 
be  lower  for  shocks  and  sudden  strains  than  for  steady  loads 
and  varying  stresses.  In  a  building  the  stresses  on  the  walls 
are  steady,  so  that  the  working  strength  may  be  taken  high  and 
hence  the  factor  of  safety  low.  In  a  bridge  the  stresses  in  the 
several  members  are  more  or  less  varying  in  character  which  re- 
quires a  lower  working  strength  and  hence  a  higher  factor  of 
safety.  In  a  machine  subject  to  shocks  the  working  strength 
should  be  lower  still  and  the  factor  of  safety  very  high. 

Twice  as  much  strain  is  theoretically  caused  by  a  suddenly 
applied  stress  as  by  one  gradually  applied.  The  complete  demon- 
stration of  this  proposition  belongs  to  the  subject  of  dynamics 
and  is  of  a  too  complex  nature  to  be  here  given.  It  is  however 
plain  that  when  the  load  or  stress  is  gradually  applied  that  it 
uniformly  increases  from  o  to  P  so  that  it  has  an  average  value 
of  y?  P  and  performs  the  work  y?  PA.  in  causing  a  given  elonga- 


ART. 


FACTORS    OF    SAFETY    AND    WORKING    STRESSES. 


tion  X,  while  the  same  work  would  be  performed  in  the  same  dis- 
tance by  the  constant  force  l/&  P.  Hence  it  might  be  supposed 
that  the  sudden  stress  P  would  produce  double  the  strain  of  the 
gradual  stress  whose  average  value  is  ^  P.  Accordingly  not  to 
impair  the  material  by  inducing  a  set  the  working  unit-stresses 
should  be  low  for  bodies  subject  to  shocks. 

The  following  are  average  values  of  the  allowable  factors  of 
safety  commonly  employed  in  American  practice.     These  values 


Material. 

For  Steady 
Stress. 

For  Varying 
Stress. 

For  Shocks. 

(Buildings.) 

(Bridges.) 

(Machines.) 

Timber 

8 

IO 

15 

Brick  &  Stone 

15 

25 

30 

Cast  Iron 

6 

10 

15 

Wrought  Iron 

4 

6 

10 

Steel 

5 

7 

10 

are  subject  to  considerable  variation  in  particular  instances,  not 
only  on  account  of  the  different  qualities  and  grades  of  the  mate- 
rial but  also  on  account  of  the  varying  judgment  of  designers. 
They  will  also  vary  with  the  range  of  varying  stress  so  that  differ- 
ent parts  of  a  bridge  may  have  very  different  factors  of  safety. 

The  proper  allowable  working  strength  of  any  material  for  ten- 
sion, compression  or  shearing,  may  be  at  once  found  by  dividing 
the  ultimate  strength  by  the  proper  factor  of  safety.  Regard  should 
also  be  paid  to  the  elastic  limit  in  selecting  the  working  strength, 
particularly  for  materials  whose  elastic  limit  is  well  defined.  For 
wrought  iron  and  steel  the  working  strength  should  be  well 
within  the  elastic  limit,  as  already  indicated  in  previous  articles. 
For  cast  iron,  stone,  brick  and  timber  it  is  often  difficult  to  deter- 
mine the  elastic  limit,  and  experience  alone  can  guide  the  proper 
selection  of  the  working  strength.  The  above  factors  of  safety 
indicate  indeed  the  conclusions  of  experiment  and  experience 
extending  over  the  past  hundred  years. 


1 8  RESISTANCE   AND   ELASTICITY   OF   MATERIALS.  I. 

The  student  should  clearly  understand  that  the  exact  values 
given  in  this  and  the  preceding  articles  would  not  be  arbitrarily 
used  in  any  particular  case  of  design.  For  instance,  if  a  given 
lot  of  wrought  iron  is  to  be  used  in  an  engineering  structure, 
specimens  of  it  should  be  tested  to  determine  its  coefficient  of 
elasticity,  elastic  limit,  ultimate  strength  and  precentage  of  elonga- 
tion. Then  the  engineer  will  decide  upon  the  proper  working 
strength,  being  governed  by  its  qualities  as  shown  by  the  tests, 
the  character  of  the  stresses  that  come  upon  it  and  the  cost  of 
workmanship. 

The  two  fundamental  principles  of  engineering  design  are 
stability  and  economy,  or  in  other  words : 

First,  the  structure  must  safely  withstand  all  the  stresses 

which  are  to  be  applied  to  it. 
Second,  the  structure  must  be  built  and  maintained  at  the 

lowest  possible  cost. 

The  second  of  these  fundamental  principles  requires  that  all  parts 
of  the  structure  should  be  of  equal  strength,  like  the  celebrated 
"  one-hoss  shay  "  of  the  poet.  For,  if  one  part  is  stronger  than 
another  it  has  an  excess  of  material  which  might  have  been  spared. 
Of  course  this  rule  is  to  be  violated  if  the  cost  of  the  labor  required 
to  save  the  material  be  greater  than  that  of  the  material  itself. 

The  factors  of  safety  stated  above  are  supposed  to  be  so 
arranged  that  if  different  materials  be  united  the  stability  of  all 
parts  of  the  structure  will  be  the  same,  so  that  if  rupture  occurs, 
everything  would  break  at  once.  Or,  in  other  words,  timber  with 
a  factor  of  safety  8  has  about  the  same  reliability  as  wrought  iron 
with  a  factor  of  4  or  stone  with  a  factor  of  15. 

The  assignment  of  working  strengths  with  regard  to  the  elastic 
limits  of  materials  is  more  rational  than  that  by  means  of  the 
factors  of  safety,  and  in  time  it  may  become  the  more  important 
and  valuable  method.  But  at  present  the  ultimate  strengths  are 
so  much  better  known  and  so  much  more  definitely  determinable 


ART.  8.     FACTORS  OF  SAFETY  AND  WORKING  STRESSES.  19 

than  the  elastic  limits  that  the  empirical  method  of  factors  of 
safety  seems  the  more  important,  due  regard  being  paid  to  con- 
siderations of  stiffness,  elastic  limit  and  ductility. 

As  an  example,  let  it  be  required  to  find  the  proper  size  of  a 
wrought  iron  rod  to  carry  a  steady  tensile  stress  of  90  OOO  pounds. 
In  the  absence  of  knowledge  regarding  the  quality  of  the  wrought 
iron  the  ultimate  strength  St  is  to  be  taken  as  the  average  value, 
5  5  ooo  pounds  per  square  inch.  Then,  for  a  factor  of  safety  of  4, 
the  working  strength  is, 

5  =  ~ =  13  750  pounds  per  square  inch. 

4 

The  area  of  cross-section  required  is  hence, 

90000        .,  - 

A  =  2 —  6.6  square  inches, 

13750 

which  may  be  supplied  by  a  rod  of  2l~  inches  diameter. 

Prob.  1 6.  Find  the  diameter  in  centimeters  of  a  wrought  iron 
rod  to  safely  carry  a  steady  stress  of  20  ooo  kilograms.  (See 
tables  in  the  Appendix.) 

Prob.  17.  A  wooden  frame  ABC  forming  an  equilateral  triangle 
consists  of  pieces  2X2  inches  jointed  at  A,  B  and  C.  It  is 
placed  in  a  vertical  plane  and  supported  at  B  and  C  so  that  BC  is 
horizontal.  Find  the  unit-stress  and  factors  of  safety  in  each  of 
the  three  pieces  when  a  load  of  4  ooo  pounds  is  applied  at  A. 

Prob.  1 8.  Determine  the  size  of  a  short  steel  piston  rod  when 
the  piston  is  15  inches  in  diameter  and  the  steam  pressure  120 
pounds  per  square  inch. 


2O  ON    PIPES,    CYLINDERS,    AND    RIVETED   JOINTS.  II. 


CHAPTER  II. 

ON  PIPES,  CYLINDERS,  AND  RIVETED  JOINTS. 

ART.  9.     WATER  AND  STEAM  PIPES. 

The  pressure  of  water  or  steam  in  a  pipe  is  exerted  in  every 
direction  and  tends  to  tear  the  pipe  apart  longitudinally.  This  is 
resisted  by  the  internal  tensile  stresses  of  the  material.  If  p  be 
the  pressure  per  square  inch  of  the  water  or  steam,  d  the  diameter 
of  the  pipe  and  /  its  length,  the  force  P  which  tends  to  cause 
longitudinal  rupture  is  p.ld.  This  is  evident  from  the  fundamental 
principle  of  hydrostatics  that  the  pressure  of  water  in  any  direc- 
tion is  equal  to  the  pressure  on  a  plane  perpendicular  to  that 
direction,  or  may  be  seen  by  imagining  the  pipe  to  be  filled  with 
a  solid  substance  on  one  side  of  the  diameter  which  would  receive 
the  pressure  p  on  each  square  inch  of  the  area  Id  and  transmit  it 
into  the  pipe.  If  t  be  the  thickness  of  the  pipe  and  5  the  work- 
ing tensile  strength  of  the  material,  the  resistance  on  each  side  is 
U.S.  As  the  resistance  must  equal  the  pressure, 

pld  =  2ttS,         or   pd  —  2tS, 
which  is  the  formula  for  discussing  pipes  under  internal  pressure. 

The  unit-pressure  p  for  water  may  be  computed  from  a  given 
head  h  by  finding  the  weight  of  a  column  of  water  one  inch  square 
and  h  inches  high.  Or  if  //  be  given  in  feet,  the  pressure  in  pounds 
per  square  inch  may  be  computed  from  p  —  0.434^. 

Water  pipes  may  be  made  of  cast  or  wrought  iron,  the  former 
being  more  common,  while  for  steam  the  latter  is  preferable. 


ART.    9.  WATER   AND   STEAM    PIPES.  21 

Wrought  iron  pipes  are  sometimes  made  of  plates  riveted  together 
but  the  discussion  of  these  is  reserved  for  another  article.  A 
water  pipe  subjected  to  the  shock  of  water  ram  needs  a  high 
factor  of  safety,  and  in  a  steam  pipe  the  factors  should  also  be  high 
owing  to  shocks  liable  to  occur  from  condensation  and  expansion 
of  the  steam.  The  formula  above  deduced  shows  that  the  thick- 
ness of  a  pipe  must  increase  directly  as  its  diameter,  the  internal 
pressure  being  constant. 

For  example,  let  it  be  required  to  find  the  factor  of  safety  for  a 
cast  iron  water  pipe  of  1  2  inches  diameter  and  ^  inches  thickness 
under  a  head  of  300  feet.  Here  p,  the  pressure  per  square  inch, 
equals  130.2  pounds.  Then  from  the  formula  the  unit-stress  is, 

pd       130  X  12 
S  =  —  -  =  -  r7~r/~  =    i  23O  pounds  per  square  inch, 

21  2   X   /8 

and  hence  the  factor  of  safety  is, 

20  ooo 

f  =  ---  =  about  ID, 
1230 

which  indicates  ample  security  against  the  shock  of  water  ram. 

Again  let  it  be  required  to  find  the  proper  thickness  for  a 
wrought  iron  steam  pipe  of  18  inches  diameter  to  resist  a  pressure 
of  1  20  pounds  per  square  inch.  With  a  factor  of  safety  of  10  the 
working  strength  S  is  about  5  500  pounds  per  square  inch.  Then 
from  the  formula, 

pd      120  X  18 

t  =     =-=°-2  mches' 


In  order  to  safely  resist  the  stresses  and  shocks  liable  to  occur  in 
handling  the  pipes,  the  thickness  is  often  made  somewhat  greater 
than  the  formula  requires. 

Prob.  19.  What  should  be  the  thickness  of  a  cast  iron  pipe  of 
1  8  inches  diameter  under  a  head  of  300  feet? 

Prob.  20.  A  wrought  iron  pipe  is  4.5  inches  in  internal  diameter 
and  weighs  12.5  pounds  per  linear  foot.  What  steam  pressure 
can  it  carry  with  a  factor  of  safety  of  8  ? 


22  ON    PIPES,    CYLINDERS,    AND    RIVETED   JOINTS.  II. 

Prob.  21.  What  head  of  water  will  burst  a  cast  iron  pipe  of  24 
inches  diameter  and        inches  thickness  ? 


ART.   10.     CYLINDERS  AND  SPHERES. 

A  cylinder  subject  to  the  internal  pressure  of  water  or  steam 
tends  to  fail  longitudinally  exactly  like  a  pipe.  The  head  of  the 
cylinder  however  undergoes  a  pressure  which  tends  to  separate  it 
from  the  walls.  If  d  be  the  diameter  of  the  cylinder  and  p  the 
internal  pressure  per  square  unit,  the  total  pressure  on  the  head  is 
^ncP.p.  If  6"  be  the  working  unit-stress  and  t  the  thickness  of 
the  cylinder,  the  resistance  to  the  pressure  is  xdt  S.  Since  the 
resistance  must  equal  the  pressure, 

y^TicP.p  =  -ndt.S,  or    pd  =  tfS. 

By  comparing  this  with  the  formula  of  the  last  article  it  is  seen 
that  the  resistance  of  a  pipe  to  tranverse  rupture  is  double  the 
resistance  to  longitudinal  rupture. 

A  sphere  subject  to  internal  pressure  tends  to  rupture  around 
a  great  circle,  and  it  is  easy  to  see  that  the  conditions  are  exactly 
the  same  as  for  the  transverse  rupture  of  a  cylinder,  or  that 
pd  =  4tS.  For  very  thick  spheres  and  cylinders  the  formulas  of 
this  and  the  last  article  are  only  approximate. 

A  cylinder  under  external  pressure  is  theoretically  in  a  similar 
condition  to  one  under  internal  pressure  as  long  as  it  remains  a 
true  circle  in  cross-section.  A  uniform  internal  pressure  tends  to 
preserve  and  maintain  the  circular  form  of  the  cylindrical  annulus, 
but  an  external  pressure  tends  at  once  to  increase  the  slightest 
variation  from  the  circle  and  render  it  elliptical.  The  distortion 
when  once  begun  rapidly  increases  and  failure  occurs  by  the  col- 
lapsing of  the  tube  rather  than  by  the  crushing  of  the  material. 
The  flues  of  a  steam  boiler  are  the  most  common  instance  of  cylin- 
ders subjected  to  external  pressure.  In  the  absence  of  a  rational 
method  of  investigating  such  cases  recourse  has  been  had  to 


ART.   10. 


CYLINDERS    AND    SPHERES. 


experiment.  Tubes  of  various  diameters,  lengths,  and  thicknesses 
have  been  subjected  to  external  pressure  until  they  collapse  and 
the  results  have  been  compared  and  discussed.  The  following 
for  instance  are  the  results  of  three  experiments  by  Fairbairn  on 
wrought  iron  tubes. 


Length 

Diameter 

Thickness 

Pressure 

in 

in 

in 

per 

Inches. 

Inches. 

Inches. 

Sq.  Inch. 

37 

9 

0.14 

378 

60 

I4tf 

O,I25 

125 

61 

i*# 

0.25 

420 

From  these  and  other  similar  experiments  it  has  been  concluded 
that  the  collapsing  pressure  varies  directly  as  some  power  of  the 
thickness,  arid  inversely  as  the  length  and  diameter  of  the  table. 
For  wrought  iron  tubes  Wood  gives  the  empirical  formula  for 
the  collapsing  pressure  per  square  inch, 


=  9  600  ooo 


Id 


The  values  of  p  computed  from  this  formula  for  the  above  three 
experiments  are  397,  120  and  409,  which  agree  well  with  the 
observed  values. 

The  proper  thickness  of  a  wrought  iron  tube  to  resist  external 
pressure  may  be  readily  found  from  this  formula  after  assuming 
a  suitable  factor  of  safety.  For  example,  let  it  be  required  to  find 
/when  p—  1  20  pounds  per  square  inch,  1=  72  inches,  d=4 
inches  and  the  factor  of  safety  =  10.  Then 


9600000 

from  which  with  the  help  of  logarithms  the  value  of  t  is  found 
to  be  0.22  inches. 


ON    PIPES,    CYLINDERS,    AND    RIVETED   JOINTS. 


II. 


Prop.  22.  What  internal  pressure  per  square  inch  will  burst  a 
cast  iron  sphere  of  24  inches  diameter  and  ^  inches  thickness. 

Prob.  23.  What  external  pressure  per  square  inch  will  collapse 
a  wrought  iron  tube  96  inches  long,  3  inches  diameter  and  0.25 
inches  thickness  ? 


ART.  ii.     RIVETED  JOINTS. 

The  strength  of  riveted  joints  which  are  subject  to  tension 
depends  upon  the  shearing  strength  of  the  rivets  and  the  tensile 
strength  of  the  plates.  Whatever  be  the  arrangement  all  parts  of 
the  joint  should  have  the  same  degree  of  stability,  so  that  at  the 
point  of  rupture  both  rivets  and  plates  (like  the  one-hoss  shay) 
may  simultaneously  fail. 

A  joint  subject  to  tension  is  always  weaker  than  the  parts 
which  it  connects,  since  a  portion  of  the  material  is  removed  to 
make  room  for  the  rivets.  It  may  be  required  to  arrange  a  joint 
so  as  to  secure  either  strength  or  tightness.  For  a  bridge,  strength 
is  mainly  needed ;  for  a  gas  holder,  tightness  is  the  principal 
requisite ;  while  for  a  boiler  both  these  qualities  are  desirable. 

Case  I.  Lap  Joint  with  single  riveting. — Let  P  be  the  tensile 
stress  which  comes  on  one  rivet,  d  the  diameter  of  a  rivet,  t  the 

thickness  of  the  plates,  and  a  the 
pitch  of  the  rivets.  Let  St  be  the 
unit  working  strength  of  the  plate 
and  S,.  that  of  the  rivet,  the  former 
being  in  tension  and  the  latter 
in  shear.  Then  for  the  plate 
P=t(a — d}  St  and  for  the  rivet 
xd*. 


P  =  '±—1  Ss.   For  equal  strength 

4 
these  must  be  equal,  or, 


ART.  ii. 


RIVETED   JOINTS. 


For  wrought  iron  plates  and  rivets  St  is  about  equal  to  J^,  and 
therefore, 


is  the  formula  for  finding  the  pitch  in  wrought  iron  lap  riveting. 

By  punching  the  hole  the  section  of  the  plate  is  reduced  from 

a.t  to  (a — d)  t,  so  that  the  ratio  of  the  strength  of  the  joint  to 

that  of  the  unriveted  plate  is, 

_  a— d  _  i 

a  l    i    4     / 

it    d 

This  shows  clearly  that  for  a  given  thickness  /,  large  rivets  give 

the  largest  value  of  r,  while  small  rivets  give  small  values  of  r. 

The  smaller  the  rivet  the  smaller  the  pitch  and  the  greater  the 

loss  in  strength.     For  example, 

ifd=f,        a  =1.78?        and     ^  =  0.44, 
\{d  =  3/.      a     \o.\t         and     ;-  — 0.77, 

Hence  when  strength  is  required  large  rivets  should  be  used, 

while   to   give   tight  joints   small    rivets  must  be  used  with  a 

sacrifice  of  strength. 

Case  II.  Lap  Joint  with  double 
riveting. — In  this  arrangement  twice 
as  many  rivets  are  used  and  hence 

P=t(a-d}St  =  2  ^~SS 
from  which,  for  wrought  iron, 

a  —  ~T7~   +  d    and  r  =  — - 


For  this  case  the  same  truth  holds  re 
garding  strength,  thus, 

when  d  =  t,  a  =  2.6t    and  r  =  0.6 1 
=  3#,  a=  17.3*  and  r=:  0.83 


Fig.  4- 


and  the  loss  of  strength  is  much  less  than  in  single  riveting. 


26  ON    PIPES,    CYLINDERS,    AND    RIVETED   JOINTS.  II. 

Case  III.  Butt  Joint  with  single  riveting. — For  this  arrangement 
the  shear  on  the  rivets  comes  on  two  cross-sections,  and  the 
^-^  covers  need  to  be  only  one- 

half  as  thick   as    the    plates. 

\f  VySSS,.  777777^*7^0-  Jj.  js  easy  J.Q  dcdUCC  for 

^r     rig.  5-  , ,  . 

wrought  iron, 
a=d-\-"- —          and       r  = 

2t  2.t 

which  is  the  same  as  in  the  preceding  case. 

Case  IV.  Butt  Joints  with  double  riveting. — Here  there  are 
two  rows  of  rivets  and  each  is  in  double  shear.  It  is  easy  to  find 
for  the  case  of  wrought  iron, 

a  =  d  -f '- —      and    r  =   


which  indicates  a  large  increase  in  strength  over  single  riveting. 

Compression  is  brought  sidewise  upon  the  rivets  in  all  the 
above  cases  by  means  of  the  stress  Pand  tends  to  cause  failure 
by  crushing.  The  exact  manner  in  which  the  compression  acts 
upon  the  cylindrical  surface  of  the  rivet  is  not  known,  but  it  is 
usually  supposed  to  be  equivalent  to  a  stress  uniformly  distrib- 
uted over  the  projection  of  the  surface  on  a  plane  through  the 
axis  of  the  rivet.  Thus  for  single  riveting  with  either  lap  or 
butt  joints, 

P  =  /#£„ 
and  for  double  riveting, 

P  =  2tdSc. 

This  compressive  or  bearing  resistance  of  rivets  always  needs  to 
be  regarded  in  actual  cases  of  design.  The  lap  of  the  plates  is 
determined  by  practical  considerations  rather  than  by  theoretic 
formulas. 


ART.    12.  MISCELLANEOUS    EXERCISES.  2/ 

For  example,  let  it  be  required  to  investigate  a  single  riveted 
butt  joint  consisting  of  plates  6  inches  wide  and  0.5  inches  thick 
with  covers  0.25  inches  thick  and  2  rivets  of  I  inch  diameter  and 
4  inches  pitch,  when  under  a  tension  of  16  ooo  pounds.  First, 
the  tensile  unit-stress  on  the  plate  is, 

S,= =  8  ooo  pounds  per  square  inch. 

4X0.5 

Next  the  shearing  unit-stress  on  the  rivets  is, 

1 6  ooo  •     1 

5  = =  10  200  pounds  per  square  inch. 

2  X  0.785 
Lastly,  the  bearing  compressive  unit-stress  on  the  rivets  is, 

5  =  —  16  ooo  pounds  per  square  inch. 

2X  i  Xo.5 

It  hence  appears  that  the  joint  has  the  greatest  resistance  against 
tension  and  the  least  against  compression. 

Prob.  24.  A  butt  joint  with  double  riveting  has  plates  half  an 
inch  thick  and  rivets  one  inch  in  diameter.  Find  the  pitch  of 
the  rivets  and  the  percentage  of  strength  lost  by  the  joint. 

Prob.  25.  A  boiler  is  to  be  formed  of  wrought  iron  plates  ^6 
inches  thick  united  by  single  lap  joints  with  rivets  %  inches  in 
diameter.  Find  the  proper  pitch  of  the  rivets.  Find  the  factor 
of  safety  of  the  boiler  if  it  is  30  inches  in  diameter  and  carries  a 
steam  pressure  of  100  pounds  per  square  inch  above  the  atmosphere. 


ART.   12.     MISCELLANEOUS  EXERCISES. 

It  will  be  profitable  to  the  student  to  thoroughly  perform  the 
following  exercises  and  to  write  upon  each  a  detailed  report 
which  should  contain  all  the  sketches  and  computations  necessary 
to  clearly  explain  the  data,  the  reasoning,  and  the  conclusions. 

Exercise  I.  Visit  an  establishment  where  tensile  tests  are 
made.  Ascertain  the  kind  of  machine  employed,  its  capacity, 
the  method  of  applying  the  stresses,  the  method  of  measuring 
the  stresses,  the  method  of  measuring  the  elongations.  Ascer- 


28  ON    PIPES,    CYLINDERS,    AND    RIVETED    JOINTS.  II. 

tain  the  kind  of  material  tested,  the  reason  for  testing  it,  and  the 
conclusions  derived  from  the  tests.  Give  full  data  for  the  tests 
on  four  different  specimens,  compute  the  values  of  coefficient  of 
elasticity,  ultimate  strength  and  ultimate  elongation  for  them, 
and  state  your  conclusions. 

Exercise  2.  Procure  a  wrought  iron  bolt  and  nut.  Determine 
diameter  of  bolt,  length  of  head,  and  length  of  nut.  State  the 
equation  of  condition  that  the  head  of  the  nut  shall  shear  off  at 
the  same  time  the  bolt  ruptures  under  tension.  Take  the  shear- 
ing strength  as  half  the  tensile  and  compute  the  length  of  head 
for  a  given  diameter.  State  why  theoretically  the  length  of  the 
nut  should  be  double  that  of  the  head.  Compare  theory  with 
practice. 

Exercise  3.  Go  to  a  boiler  shop  and  witness  operations  upon  a 
boiler  in  process  of  construction.  Ascertain  length  and  diameter 
of  boiler,  thickness,  pitch  and  diameter  of  rivets,  method  of 
forming  holes,  method  of  doing  the  riveting.  Compute  the  loss 
of  strength  caused  by  the  riveting.  Compute  the  steam  pressure 
which  would  cause  longitudinal  rupture  of  the  plate  along  a 
line  of  rivets.  Ascertain  whether  the  joint  is  proportioned  in 
accordance  with  theory. 

Prob.  26.  A  bar  whose  cross-section  is  A  is  subjected  to  a 

tensile  stress  P.     Prove  that  a  shear  exists  along  any  oblique 

p 

section  and  that  the  maximum  shearing  unit-stress  is  l/2  — . 

A 
Prob.  27.  A  wrought  iron  pipe  I/Q  inches  thick  and  20  inches 

in  diameter  is  to  be  subjected  to  a  head  of  water  of  230  feet. 

Compute  the  probable  increase  in  diameter  due  to   the  internal 

pressure. 


ART.  13.  DEFINITIONS.  29 


CHAPTER   III. 

ON  SIMPLE  BEAMS  AND   CANTILEVERS. 

ART.    13.     DEFINITIONS. 

Transverse  stress,  or  flexure,  occurs  when  a  bar  is  laid  in  a 
horizontal  position  upon  one  or  more  supports.  The  weight  of 
the  bar  and  the  loads  upon  it  cause  it  to  bend  and  induce  in  it 
stresses  and  strains  of  a  complex  nature  which,  as  will  be  seen 
later,  may  be  resolved  into  those  of  tension,  compression,  and 
shear.  Such  a  bar  is  called  a  beam. 

A  simple  beam  is  a  bar  resting  upon  supports  at  its  ends.  A 
cantilever  beam  is  a  bar  on  one  support  in  its  middle,  or  the 
portion  of  any  beam  projecting  out  of  a  wall  or  beyond  a  support 
may  be  called  a  cantilever.  A  continuous  beam  is  a  bar  resting 
upon  more  than  two  supports.  In  this  book  the  word  beam, 
when  used  without  qualification,  includes  all  kinds,  whatever  be 
the  number  of  the  supports  or  whether  the  ends  be  free,  supported, 
or  fixed. 

The  elastic  curve  is  the  curve  formed  by  a  beam  as  it  deflects 
downward  under  the  action  of  its  own  weight  and  of  the  loads 
upon  it.  Experience  teaches  that  the  amount  of  this  deflection 
and  curvature  is  very  small.  A  beam  is  said  to  be  fixed  at  one  end 
when  it  is  so  arranged  that  the  tangent  to  the  elastic  curve  at 
that  end  always  remains  horizontal.  This  may  be  done  in  prac- 
tice by  firmly  building  one  end  into  a  wall.  A  beam  fixed  at  one 
end  and  unsupported  at  the  other  is  a  cantilever. 


3O  ON    SIMPLE    BEAMS    AND    CANTILEVERS.  III. 

The  loads  on  beams  are  either  uniform  or  concentrated.  A 
uniform  load  embraces  the  weight  of  the  beam  itself  and  any 
load  evenly  spread  over  it.  Uniform  loads  are  estimated  by  their 
intensity  per  unit  of  length  of  the  beam,  and  usually  in  pounds 
per  linear  foot.  The  uniform  load  per  linear  unit  is  designated 
by  w,  then  wx  will  represent  the  load  over  any  distance  x.  If  / 
be  the  length  of  the  beam  the  total  uniform  load  is  wl  which 
may  be  represented  by  W.  A  concentrated  load  is  a  weight 
applied  at  a  definite  point  and  is  designated  by  P. 

In  this  chapter  cantilevers  and  simple  beams  will  be  principally 
discussed,  although  all  the  fundamental  principles  and  methods 
hold  good  for  restrained  and  continuous  beams  as  well.  Unless 
otherwise  stated  the  beams  will  be  regarded  as  of  uniform  cross- 
section  throughout,  and  in  computing  their  weights  the  rules  of 
Art.  i  will  be  found  of  service. 

Prob.  28.  Find  the  diameter  of  a  round  steel  bar  which  weighs 
12  pounds  per  linear  foot. 

ART.  14.     REACTIONS  OF  THE  SUPPORTS. 

The  points  upon  which  a  beam  is  supported  react  upward 
against  the  beam  an  amount  equal  to  the  pressure  of  the  beam 
upon  them.  The  beam,  being  at  rest,  is  a  body  in  equilibrium 
under  the  action  of  a  system  of  forces  which  consist  of  the  down- 
ward loads  and  the  upward  reactions.  The  loads  are  usually 
given  in  intensity  and  position  and  it  is  required  to  find  the  re- 
actions. This  is  effected  by  the  application  of  the  fundamental 
conditions  of  static  equilibrium,  which  for  a  system  of  vertical 
forces,  are, 

2  of  all  vertical  forces  =  o, 

2'  of  moments  of  all  forces  =  o. 

The  first  of  these  conditions  says  that  the  sum  of  all  the  loads 
on  the  beam  is  equal  to  the  sum  of  the  reactions.  Hence  if  there 
be  but  one  support,  the  condition  gives  at  once  the  reaction. 


ART.    14.  REACTIONS   OF  THE   SUPPORTS.  3! 

For  two  supports  the  second  condition  must  be  used  in   con- 
nection with  the  first.     The  center  of  moments  may  be  taken 
anywhere  in  the  plane,  but  it  is  more  convenient  to  take  it  at  one 
of  the  supports.     For  example,  consider  a  single  concentrated 
load  P  situated  at  4  feet  from 
the  left  end  of  a  simple  beam 
whose  span  is    13  feet.     The 

equation    of  moments,    with  Rl  Tig. 6.  "fl, 

the  center  at  the  left  support, 

is  13  R2  —  4  P=  o,  from  which  R9  =  ±P.  Again  the  equa- 
tion of  moments,  with  the  center  at  the  right  support,  is 
1 3  Rt  —  9  P  =  o,  from  which  RT  —  ^P.  As  a  check  it  may  be 
observed  that  RT  +  R9  =  P. 

For  a  uniform  load  over  a  simple  beam  it  is  evident,  without 
applying  the  conditions  of  equlibrium,  that  each  reaction  is  one- 
half  the  load. 

For  an  overhanging  beam  uniformly  loaded,  as  shown  in  Fig.  7, 
let  w  be  the  load  per  linear 

unit.  For  a  center  of  moments          i  y—m--^ 

at  the  right  support   the  sec-      ' — 5 s ' 

ond  condition  gives,  Ij^  Hff-T-  *H» 

T->   i  t  f    \  m 

R  ./  —  wl.  —  -f-  wm.  —  =  o, 

2  2 

while  with  the  center  at  the  left  support, 

R  ./  —  wl  —  —    wm  (I  -\- )  =  o. 

2  2 

From  these  it  is  easy  to  deduce  the  reactions, 

wm2                            wl  wm2 

R    = h  wm  4-  — -r. 


2l   •  '2  2l' 

whose  sum  is  equal  to  the  total  load  wl  -\-  wm.  Here,  as  in  all 
cases  of  uniform  load,  the  lever  arms  are  taken  to  the  centers  of 
gravity  of  the  portion  considered. 


32 


ON    SIMPLE    BEAMS   AND    CANTILEVERS. 


III. 


When  there  are  more  than  two  supports  the  problem  of  rind- 
ing the  reactions  from  the  principles  of  statics  becomes  indetermi- 
nate, since  two  conditions  of  equilibrium  are  only  sufficient  to 
determine  two  unknown  quantities.  By  introducing,  however, 
the  elastic  properties  of  the  material,  the  reactions  of  continuous 
beams  may  be  deduced  as  will  be  explained  in  the  next  chapter. 

Prob.  29.  Three  men  carry  a  stick  of  timber,  one  taking  hold 
at  one  end  and  the  other  two  at  a  common  point.  Where  should 
this  point  be  so  that  each  may  bear  one-third  of  the  weight?  -  '/ 

Prob.  30.  A  simple  beam  weighing  30  pounds  per  linear  foot 
is  1  8  feet  long.  A  weight  of  700  pounds  is  placed  5  feet  from 
the  left  end  and  one  of  500  pounds  at  10  feet  from  the  left  end. 
Find  the  reactions  due  to  the  total  load. 


ART.  15.     EXTERNAL  FORCES  AND  INTERNAL  STRESSES. 

The  external  loads  and  reactions  on  a  beam  maintain  their 
equilibrium  by  means  of  internal  stresses  which  are  generated  in 
it  It  is  required  to  determine  the  relations  between  the  external 
forces  and  the  internal  stresses. 


II 


Consider  a  beam  of  any  kind 
loaded  in  any  manner.  Imagine 
a  plane  mn  cutting  the  beam  at 
any  cross-section.  In  that  section 
there  are  acting  unknown  stresses 
of  various  intensities  and  direc- 
tions. Let  the  beam  be  imagined 
to  be  separated  into  two  parts  by 
the  cutting  plane  and  let  forces  X, 
Y,  Z,  etc.,  equivalent  to  the  internal 
stresses,  be  applied  to  the  section 
as  shown  in  Fig.  8.  Then  the 

equilibrium  of  each  part  of  the  beam  will  be  undisturbed,  for  each 
part  will  be  acted  upon  by  a  system  of  forces  in  equilibrium. 


it 


ART.    15.      EXTERNAL   FORCES    AND    INTERNAL   STRESSES.  33 

Hence  the  following  fundamental  principle  is  established, 

The  internal  stresses  in  any  cross-section  of  a  beam  hold  in 
equilibrium  the  external  forces  on  each  side  of  that  section. 

This  is  the  most  important  principle  in  the  theory  of  flexure. 
It  applies  to  all  beams,  whether  the  cross-section  be  uniform  or 
variable  and  whatever  be  the  number  of  the  spans  or  the  nature 
of  the  loading. 

Thus  in  the  above  figure  the  internal  stresses  X,  Y,  Z,  etc.,  hold 
in  equilibrium  the  loads  and  reactions  on  the  left  of  the  section, 
and  also  those  on  the  right.  Considering  one  part  only  a  system 
offerees  in  equilibrium  is  seen,  to  which  the  three  necessary  and 
sufficient  conditions  of  statics  apply,  namely, 

2  of  all  horizontal  components  =  o, 
2  of  all  vertical  components  =  o, 
I  of  moments  of  all  forces  =  o. 

From  these  conditions  can  be  deduced  three  laws  concerning 
the  unknown  stresses  in  any  section.  Whatever  be  the  intensity 
and  direction  of  these  stresses,  let  each  be  resolved  into  its  hori- 
zontal and  vertical  components.  The  vertical  components  will 
add  together  and  form  a  certain  resultant  force 

V  which  tends  to  shear  off  the  section  from  the     ^t 

one  adjacent  to  it.    The  horizontal  components 
will  be  applied  at  different  points  of  the  cross- 


section,  some  acting  in  one  direction  and  some 
in  the  other,  or  in  other  words,  some  of  the  horizontal  stresses  are 
tensile  and  some  compressive.    Hence  for  any  section  of  any  beam 
the  following  laws  concerning  the  internal  stresses  may  be  stated. 

ist.  The  algebraic  sum  of  the  horizontal  stresses  is  zero;  or 
the  sum  of  the  horizontal  tensile  stresses  is  equal  to  the 
sum  of  the  horizontal  compressive  stresses. 

2nd.  The  algebraic  sum  of  the  vertical  stresses  forms  a 
resultant  shear  which  is  equal  to  the  algebraic  sum  of  the 
external  vertical  forces  on  either  side  of  the  section. 


34  ON   SIMPLE   BEAMS   AND   CANTILEVERS.  III. 

3rd.  The  algebraic  sum  of  the  moments  of  the  internal 
stresses  is  equal  to  the  algebraic  sum  of  the  moments  of 
the  external  forces  on  either  side  of  the  section. 

These  three  theoretical  laws  are  the  foundation  of  the  theory 
of  the  flexure  of  beams.  Their  expression  may  be  abbreviated 
by  introducing  the  following  definitions. 

'  Resisting  shear '  is  the  name  given  to  the  algebraic  sum  of 
the  vertical  stresses  in  any  section,  and  '  vertical  shear '  is  the 
name  for  the  algebraic  sum  of  the  external  vertical  forces  on  the 
left  of  the  section.  '  Resisting  moment '  is  the  name  given  to 
the  algebraic  sum  of  the  moments  of  the  internal  horizontal 
stresses  with  reference  to  a  point  in  the  section,  and  'bending 
moment'  is  the  name  for  the  algebraic  sum  of  the  moments  of 
the  external  forces  on  the  left  of  the  section  with  reference  to  the 
same  point.  Then  the  three  laws  may  be  thus  expressed  for  any 
section  of  any  beam, 

Sum  of  tensile  stresses  =  Sum  of  compressive  stresses. 

Resisting  shear  =  Vertical  shear. 
Resisting  moment  =  Bending  moment. 

The  second  and  third  of  these  equations  furnish  the  fundamental 
formulas  for  investigating  beams.  They  state  the  relations  be- 
tween the  internal  stresses  in  any  section  and  the  external  forces 
on  the  left-hand  side  of  that  section. 

Prob.  31.  A  wooden  beam  12  X  14  inches  and  6  feet  long  is 
supported  at  one  end  by  a  force  of  560  pounds  acting  at  an  angle 
of  60  degrees  with  the  vertical,  and  at  the  other  end  by  a  vertical 
force  Fand  a  horizontal  force  X.  Find  the  values  of  X  and  Y. 

ART.  1 6.     THE  VERTICAL  SHEAR. 

Vertical  Shear  is  the  name  given  to  the  algebraic  sum  of  all 
external  forces  on  the  left  of  the  section  considered.  Let  it  be 
denoted  by  V,  then  for  any  section, 

V  •=•  Reactions  on  left  of  section  minus  all  loads  on  left. 


ART.    1 6.  THE    VERTICAL   SHEAR.  35 

Here  upward  forces  are  regarded  as  positive  and  downward 
forces  as  negative.  V  is  hence  positive  or  negative  accord- 
ing as  the  reaction  exceeds  or  is  less  than  the  loads  on  the  left 
of  the  section.  To  illustrate,  consider  a  simple  beam  or  canti- 
lever loaded  in  any  manner  and 

cut  at  any  section  by  a  vertical  ip  Fm          .      f** 

plane  mn.     Let  R  be  the  left  and     |   J  ^     ^  \ 

R'  the  right  reaction.     Let   IP    H5  ~~  "" 

denote  the  sum  of  all  the  loads         ** 
on  the  left  of  the  section  and  IP 
the  sum  of  those  on  the  right.     Then,  from  the  definition, 

V=R-IP. 

Since  R  +  R'  =  IP  +  IP'  it  is  clear  if  R  —  IP  =  +  Fthat 
R'  —  IP'  =  —  V,  or  that  the  resultant  of  all  the  external  forces 
on  one  side  of  the  section  is  equal  and  opposite  to  the  resultant 
of  those  on  the  other  side.  They  form,  in  short,  a  pair  of  shears 
acting  very  near  together  on  either  side  of  the  section  and  tend- 
ing to  cause  a  sliding  or  detrusion  along  the  section.  The  value 
of  the  vertical  shear  for  any  section  of  a  simple  beam  or  canti- 
lever is  readily  found  by  the  above  equation.  When  R  exceeds 
IP,  the  vertical  shear  V  is  positive,  and  the  left  part  of  the  beam 
tends  to  slide  upward  relative  to  the  right  part.  When  R  is  less 
than  IP,  the  vertical  shear  V  is  negative,  and  the  left  part  tends 
to  slide  downward  relative  to  the  other. 

The  vertical  shear  varies  greatly  in  value  at  different  sections 
of  a  beam.  Consider  first  a  simple  beam  /  feet  long  and  weigh- 
ing w  pounds  per  linear  foot.  Each  reaction  is  then  ^wl.-  Pass 
a  plane  at  any  distance  x  from  the  left  support,  then  from  the 
definition  the  vertical  shear  for  that  section  is  V  =  y^wl  —  wx. 
Here  it  is  seen  that  V  has  its  greatest  value  y^wl  when  x  =  o, 
that  V  decreases  as  x  increases,  and  that  V  becomes  o  when 
x  =  y^l.  When  x  is  greater  than  y2l,  V is  negative  and  becomes 
when  x=l.  The  equation  V  =  y^wl  —  wx  is  indeed 


ON    SIMPLE    BEAMS   AND    CANTILEVERS. 


III. 


the  equation  of  a  straight  line,  the  origin  being  at  the  left  support, 

and  may  be  plotted  so  that  the 

r~ -j 1  ordinate  at  any  point  of  the  beam 

"^  will  represent  the  vertical  shear 
for  that  point,  as  shown  in  Fig.  1 1. 


Consider  again  a  simple  beam 
loaded  with  several  weights  PIt 

P2,  P3,  etc.,  and  let  the  weight  of  the  beam  itself  be  neglected. 

Here  for  any  point  a,  between  the  support  and  the  first  load, 

V=R\  for  b,  V=R  —  P1\  for 
c,  V  =  R  —  PT  —  P2,  and  so  on. 
For  the  case  of  three  loads  the 
graphical  representation  of  verti- 
cal shears  is  shown  in  Fig.  12. 

For  a  cantilever  beam  there  is 
no  reaction  at  the  left  end,  and  for 

any  section,  V  =  —  -P.     In  any  case  2P  must  include  both 

uniform  and  concentrated  loads  if  such  are  upon  the  beam.  For 
,  a  cantilever  with  two  concen- 
trated loads  and  a  uniformly 
distributed  load  the  vertical 
shear  for  a  is  V  =  — PT  —  wx, 
and  for  6,  V=  —P,— P2—  wx, 
where  x  is  the  distance  from 
the  left-hand  end,  and  the 
graphical  representation  is 

shown  in  Fig.  13.     Here  Vis  a  maximum  at  the  wall  and  then 

immediately  becomes  o. 

Prob.  32.  A  simple  beam  12  feet  long  and  weighing  20  pounds 
per  linear  foot  has  a  load  of  600  pounds  at  2  feet  from  the  left 
end.  Find  the  vertical  shears  at  the  ends  and  under  the  load. 
Draw  a  diagram  to  show  the  distribution  of  vertical  shears. 


I  ,','", 


ART.    17.  THE    BENDING    MOMENT.  37 

ART.   17.     THE  BENDING  MOMENT. 

Consider  again  any  beam  cut  at  any  section  by  a  vertical  plane. 
The  third  fundamental  condition  of  static  equilibrium  (Art.  15) 
states  that  the  algebraic  sum  of  the  moments  of  the  external 
forces  on  one  side  of  the  section  is  equal  to  the  sum  of  the  resist- 
ing moments  of  the  internal  stresses  in  the  section.  This  is  true 
for  any  center  of  moments. 

Bending  moment  is  the  name  given  to  the  algebraic  sum  of 
the  moments  of  the  external  forces  on  the  left  of  the  section  with 
reference  to  a  point  in  that  section.  Let  it  be  denoted  by  M. 
Then,  for  a  cantilever  or  simple  beam, 

M  =  moment  of  reaction  minus  sum  of  moments  of  loads. 
Here  the  moment  of  upward  forces  is  taken  as  positive  and  that 
of  downward  forces  as  negative.     M  may  hence  be  positive  or 
negative  according  as  the  first  or  second  term  is  the  greater. 

For  a  simple  beam  of  length  /,  uniformly  loaded,  each  reaction  is 
y^wl.  For  any  section  distant  x 

from  the  left  support  the  bending         {*      ^  -__j 

moment  is M=  ^AwLx — wx.lAx,        <£  ' """^ 


! 


or  M  =  y2w  (Ix  —  x2).  Here 
M=  o  when  x=  o  and  also  when 
x  =  I,  and  Mis  a  maximum  when 

x  =  y2l.    The  equation,  in  short, 

wl2 
is  that  of  a  parabola  whose  maximum  ordinate  is  __   and  whose 

8 

graphical  representation  is  as  given  in  Fig.  14,  each  ordinate 
showing  the  value  of  M  for  the  corresponding  value  of 
the  abscissa  x. 

Consider  next  a  simple  beam  loaded  with  only  three  weights 
PIf  P2  and  P3.  Here  for  any  point  between  the  left  support  and 
the  first  load  M=  Rx,  and  for  any  point  between  the  first  and 
second  loads  M=Rx  —  P^(x — a).  Each  of  these  expressions  is 
the  equation  of  a  straight  line,  x  being  the  abscissa  and  M  the 


210973 


38  ON    SIMPLE    BEAMS    AND    CANTILEVERS.  III. 

ordinate,  and  it  is  easy  to  see  that  the  graphical  representation  of 

bending  moments    is   as    shown 
IE     IB  rS  „. 

rH      1 ]  m  Fig-  15- 

For  a  cantilever  there  is  no 
reaction  at  the  left  end  and  all 
the  bending  moments  are  nega- 
tive. For  instance,  for  a  cantilever 

uniformly  loaded  and  having  a  load  at  the  end  the  bending  mo- 
ment is  M  =  —  Px  —  y2wxz.  Here  the  variation  of  moments 
may  be  represented  by  a  parabola,  M  being  o  at  the  free  end 
and  a  maximum  at  the  wall. 

For  any  given  case  the  bending  moment  at  any  section  may 
be  readily  found  by  using  the  definition  given  above.  The  bend- 
ing moment  in  all  cases  is  a  measure  of  the  tendency  of  the  ex- 
ternal forces  on  the  left  of  the  section  to  turn  the  beam  around  a 
point  in  that  section.  This  turning  is  prevented  by  the  internal 
resisting  moments  of  the  stresses  in  the  section,  whose  sum 
exactly  equals  the  bending  moment. 

The  bending  moment  is  a  compound  quantity  resulting  from 
the  multiplication  of  a  force  by  a  distance.  Usually  the  forces 
are  expressed  in  pounds  and  the  distances  in  feet  or  inches ;  then 
the  bending  moments  are  pound-feet  or  pound-inches.  Thus  if 
a  load  of  $00  pounds  be  at  the  middle  of  a  simple  beam  of  8 
feet  span,  the  bending  moment  under  the  load  is, 

M  =  250  X  4  =  i  ooo  pound-feet  =  12  ooo  pound-inches. 
Again  let  a  simple  beam  of  8  feet  span  be  uniformly  loaded  with 
500  pounds.     Then  the  bending  moment  at  the  middle  is, 
M  =  250  X  4  —  250  X  2  =  500  pound-feet  =  6  ooo 

pound-inches. 

Prob.  33.  A  simple  beam  of  12  feet  span  weighs  20  pounds 
per  linear  foot  and  has  a  load  of  250  pounds  at  3  feet  from  the 
left  end.  Find  the  bending  moments  for  the  quarter  points  and 
for  the  middle  of  the  beam. 


ART.    1 8.         THEORETICAL   AND    EXPERIMENTAL    LAWS.  39 

Prob.  34.  A  beam  6  feet  long  and  weighing  20  pounds  per 
foot  is  placed  upon  a  single  support  at  its  center.  Compute  the 
bending  moments  for  sections  distant  I,  2,  3  and  4  feet  from  the 
left  end,  and  draw  a  curve  to  show  the  distribution  of  moments 
throughout  the  beam. 

ART.  1 8.  THEORETICAL  AND  EXPERIMENTAL  LAWS. 
From  the  three  necessary  conditions  of  static  equilibrium,  as 
stated  in  Art.  15,  three  important  theoretical  laws  regarding  in- 
ternal stresses  were  deduced.  These  alone,  however,  are  not 
sufficient  for  the  full  investigation  of  the  subject,  but  recourse 
must  be  had  to  experience  and  experiment.  Experience  teaches 
that  when  a  beam  deflects  one  side  becomes  concave  and  the 
other  convex,  and  it  is  reasonable  to  suppose  that  the  horizontal 
tensile  stresses  are  on  the  convex  side  and  the  compressive  stresses 
on  the  concave.  By  experiments  on  beams  this  is  confirmed  and 
the  following  laws  deduced. 

(F) — The  horizontal  fibers  on  the  convex  side  are  elongated 
and  those  on  the  concave  are  shortened,  while  near  the 
center  is  a  neutral  surface  which  is  unchanged  in  length. 

(G) — The  amount  of  elongation  or  compression  of  any  fiber 
is  directly  proportional  to  its  distance  from  the  neutral 
axis.  Hence  by  law  (B)  the  horizontal  stresses  are  also 
directly  proportional  to  their  distances  from  the  neutral 
axis,  provided  the  elastic  limit  of  the  material  be  not 
exceeded. 

The  neutral  surface  passes  through  the  centers  of  gravity  of 
the  cross-sections.  To  prove  this  let  a  be  the  area  of  any  ele- 
mentary fiber  and  z  its  distance  from  the  neutral  surface.  Let  S 
be  the  unit-stress  on  the  fiber  most  remote  from  the  neutral  sur- 
face at  the  distance  c.  Then  by  law  (G), 

—  =±=  unit-stress  at  the  distance  unity, 
c 

—  z  —  unit-stress  at  the  distance  z, 


4O  ON    SIMPLE    BEAMS   AND    CANTILEVERS.  III. 

therefore  —  az  =  the  total  stress  on  any  fiber  of  area  a, 

and    -F =  algebraic  sum  of  all  horizontal  stresses. 

But  by  the  first  law  of  Art.  1 5  this  algebraic  sum  is  zero,  and 
since  S  and  c  are  constants  Saz  =  o.  This  however  is  the  con- 
dition which  makes  the  line  of  reference  pass  through  the  center 
of  gravity  as  shown  in  elementary  mechanics.  Therefore  the 

neutral  surface  of  beams  passes 
through  the  centers  of  gravity 
of  the  cross-section. 

The  '  neutral  axis '  of  a  cross- 
section  is  the  line  in  which  the 

Fiei6.  "  neutral  surface   intersects   the 

plane  of  the  cross-section.     On 

the  left  of  Fig.  16  is  shown  the  neutral  axis  of  a  cross-section 
and  on  the  right  a  trace  of  the  neutral  surface. 

Prob.  35.  A  beam  3  inches  wide  and  6  inches  deep  is  strained 
so  that  the  unit-stress  at  the  remotest  fiber  of  a  certain  cross- 
section  is  600  pounds  per  square  inch.  Find  the  sum  of  all  the 
tensile  stresses  on  the  cross-section. 

ART.  19.     THE  RESISTING  SHEAR  AND  THE  RESISTING 

MOMENT. 

The  resisting  shear  is  the  algebraic  sum  of  all  the  vertical 
components  of  the  internal  stresses  at  any  section  of  the  beam. 
If  A  be  the  area  of  that  section  and  5^  the  shearing  unit-stress, 
regarded  as  uniform  over  the  area,  then  from  formula  (i), 

Resisting  shear  =  ASS. 

The  resisting  moment  is  the  algebraic  sum  of  the  moments  of 
the  internal  horizontal  stresses  at  any  section  with  reference  to 
a  point  in  that  section.  To  find  an  expression  for  its  value  let  5 
be  the  horizontal  unit-stress,  tensile  or  compressive  as  the  case 


ART.    20.  THE   TWO    FUNDAMENTAL    FORMULAS.  4! 

may  be,  upon  the  fiber  most  remote  from  the  neutral  axis  and  let 
c  be  the  shortest  distance  from  that  fiber  to  said  axis.  Also 
let  z  be  the  distance  from  the  neutral  axis  to  any  fiber  having  the 
elementary  area  a.  Then  by  law  (£)  and  Fig.  16, 

—  =  unit-stress  at  a  distance  unity, 
c 

_  z  =  unit-stress  at  distance  z, 
c 

2 
c 

and  .  ——  =  moment  of  this  stress  about  neutral  axis. 
c 

;.    2 —  =  resisting  moment  of  horizontal  stresses. 

c 
Since  5  and  c  are  constants  this    expression    may  be  written 

—  laz2.      But  2az2,  being  the  sum  of  the  products  formed  by  mul- 

c 

tiplying  each  elementary  area  by  the  square  of  its  distance  from 

the  neutral  axis,  is  the  moment  of  inertia  of  the  cross-section 
with  reference  to  that  axis  and  may  be  denoted  by  /.  Therefore, 

SI 
Resisting  moment  =  — . 

Prob.  36.  A  wooden  beam  6X12  inches  and  five  feet  long  is 
supported  at  one  end  and  kept  level  by  two  horizontal  forces  X 
and  Y  acting  at  the  other  end  in  the  median  line  of  the  cross- 
section,  the  former  at  2  inches  from  the  top  and  the  latter  at  2 
inches  from  the  base.  Find  the  values  of  X  and  Y. 


ART.  20.     THE  Two  FUNDAMENTAL  FORMULAS. 

Consider  again  any  beam  loaded  in  any  manner  and  cut  at  any 
section  by  a  vertical  plane.  The  internal  stresses  in  that  section 
hold  in  equilibrium  the  external  forces  on  the  left  of  the  section, 
and  as  shown  in  Art.  15, 

Resisting  shear  =  Vertical  shear, 
Resisting  moment  i=  Bending  moment. 


42  ON    SIMPLE    BEAMS    AND    CANTILEVERS.  III. 

In  the  last  article  values  of  the  resisting  shear  and  the  resisting 
moments  were  deduced.  These  two  equations  then  become, 

(3)  W=r, 

(4)  S-j-  =  M, 

which  are  the  fundamental  formulas  for  investigating  the  strength 
of  beams.  From  (3)  the  average  shearing  unit-stress  5^  at  any 
cross-section  may  be  found.  From  (4)  the  tensile  or  compressive 
unit-stress  5  upon  the  remotest  fiber  at  any  cross-section  may  be 
computed.  It  is  now  seen  that  transverse  stresses  are  investigated 
by  resolving  them  into  the  simple  stresses  of  tension,  compression 
and  shear.  Whether  S  be  tension  or  compression  depends,  in 
any  particular  case,  upon  whether  c  is  measured  to  the  concave 
or  convex  side  of  the  beam. 

Fis  readily  found,  as  explained  in  Art.  16,  for  any  given  section 
of  any  beam.  Usually  only  its  maximum  values  are  needed  in 
investigations  of  strength  and  these  will  be  found  at  the  supports. 

M  is  readily  found,  as  explained  in  Art.  17,  for  any  given  sec- 
tion. Usually  only  its  maximum  values  need  be  determined  and 
these  are  near  the  middle  for  a  simple  beam  and  at  the  support 
for  a  cantilever.  The  values  of  c  and  /  for  any  given  cross-section 
must  be  computed  by  the  well-known  methods  explained  in  ele- 
mentary mechanics.  Then  the  unit-stress  S  will  be  found  from  (4). 

Experience  and  experiment  teach  us  that  simple  beams  of  uni- 
form section  break  near  the  middle  by  the  tearing  or  crushing  of 
the  fibers  and  very  rarely  at  the  supports  by  shearing.  Hence 
it  is  formula  (4)  that  is  mainly  needed  in  the  practical  investigation 
of  beams.  The  following  example  and  problem  relate  to  formula 
(3)  only,  while  formula  (4)  will  receive  detailed  discussion  in  the 
subsequent  articles. 

As  an  example,  consider  a  wrought  iron  I  beam  1 5  feet  long 
and  weighing  200  pounds  per  yard,  over  which  roll  two  locomo- 
tive wheels  6  feet  apart  and  each  bearing  12  ooo  pounds.  The 


ART.    21.  CENTER   OF   GRAVITY    OF    CROSS-SECTIONS.  43 

maximum  vertical  shear  at  the  left  support  will  evidently  occur 
when  one  wheel  is  at  the  support.  The  reaction  will  then  be 
500  +  12  ooo  -f-  ^  X  12  ooo  =  19  700  pounds.  The  greatest 
value  of  Fin  the  beam  is  then  19  700  pounds.  As  the  area  of 
the  cross-section  is  20  square  inches  the  average  shearing  unit- 
stress  is  985  pounds  so  that  the  factor  of  safety  is  about  50. 

Prob.  37.  A  wooden  beam  6X9  inches  and  12  feet  in  span 
carries  a  uniform  load  of  20  pounds  per  foot  besides  its  own 
weight  and  also  two  wagon  wheels  one  weighing  4  ooo  pounds 
and  the  other  3  ooo  pounds.  Find  the  factor  of  safety  against 
shearing. 

ART.  21.     CENTER  OF  GRAVITY  OF  CROSS-SECTIONS. 

The  fundamental  formula  (4)  contains  c,  the  shortest  distance 
from  the  remotest  part  of  the  cross-section  to  a  horizontal  axis 
passing  through  the  center  of  gravity  of  that  cross-section.  The 
methods  of  finding  c  are  explained  in  books  on  theoretical 
mechanics  and  will  not  here  be  repeated.  Its  values  for  some  of 
the  simplest  cases  are  however  recorded  for  reference. 

For  a  rectangle  whose  height  is  d,  c  =  y2d. 

For  a  circle  whose  diameter  is  d,  c  =  yzd. 

For  a  triangle  whose  altitude  is  d,  c  =  2/^d. 

For  a  square  with  side  d  having  one   diagonal 

vertical,  c  =  d\/^. 

For  a  I  whose  depth  is  d,  c  =  y%d. 

For  a  1  whose  depth  is  d,  thickness  of  flange  t, 
width  of  flange  b,  and  thickness  of  web  /' 


t'd-\-  t(b  —  t') 
For  a  trapezoid  whose  depth  is  d,  upper  base  b, 

and  lower  base  V  ',  c  =  b  +  2  b—  - 

b  +  V  3- 

The  student  should  be  prepared  to  readily  apply  the  principle  of 
moments  to  the  deduction  of  the  numerical  value  of  c  for  any 


)=&-*        -K 

p  '#.-  i 


?4/3  *.  B> 

Q      .  -  '   -v    ^  ^     -  i. ^  •  ^ 

•=  fA  C  - 


44  ON   SIMPLE    BEAMS    AND    CANTILEVERS.  III. 

given  cross-section.  In  nearly  all  cases  the  given  area  may  be 
divided  into  rectangles,  triangles,  and  circular  areas,  whose  centers 
of  gravity  are  known,  so  that  the  statement  of  the  equation  for 
finding  c  is  very  simple. 

Prob.  38.  Find  the  value  of  c  for  a  rail  headed  beam  whose 
section  is  made  up  of  a  rectangular  flange  ^  X  4  inches,  a  rectan- 
gular web  y^  X  5  inches,  and  an  elliptical  head  ^  inches  deep 
and  i  YZ  inches  wide. 

ART.  22.     MOMENT  OF  INERTIA  OF  CROSS-SECTIONS. 

The  fundamental  formula  (4)  contains  7,  the  moment  of  inertia 
of  the  cross-section  of  the  beam  with  reference  to  a  horizontal 
axis  passing  through  the  center  of  gravity  of  that  cross-section. 
Methods  of  determining  /  are  explained  in  works  on  elementary 
mechanics  and  will  not  here  be  repeated,  but  the  values  of  some 
of  the  most  important  cases  are  recorded  for  reference. 

For  a  rectangle  of  base  b  and  depth  d,  I  =  —  . 

For  a  circle  of  diameter  d*  I  =  -  __ 

64 
For  an  ellipse  with  axes  a  and  b,  the  latter 

vertical, 


For  a  triangle  of  base  b  and  depth  d,  I  =  -- 


64 
I  = 
For  a  square  with  side  d  having  one  diag- 

onal vertical,  7  =  -- 

12 

For  a  I  with  base  b,  depth  d,  thickness  of 
flanges  t  and  thickness  of  web  ^, 


12 

For  an  1  with  base  b,  depth  d,  thickness 
of  flange  t,  thickness  of  web  f  and 
area  A,  I  =  gi=tf-/0(rf-fr  _  Ac,, 


ART.    23.  THE    MAXIMUM    BENDING   MOMENT.  45 

The  value  of  /  for  any  given  section  may  always  be  computed 
by  dividing  the  figure  into  parts  whose  moments  of  inertia  are 
known  and  transferring  these  to  the  neutral  axis  by  means  of  the 
familiar  rule  /z  =  70  -f-  Ah2,  where  70  is  the  primitive  value,  /r  the 
value  for  any  parallel  axis,  A  the  area  of  the  figure  and  h  the  dis- 
tance between  the  two  axes. 

Prob.  39.  Find  the  moment  of  inertia  of  a  triangle  with  refer- 
ence to  its  base,  and  also  with  reference  to  a  parallel  axis  passing 
through  its  vertex. 

Prob.  40.  Compute  the  least  moment  of  inertia  of  a  trapezoid 
whose  altitude  is  3  inches,  upper  base  2  inches  and  lower  base  5 
inches. 

ART.  23.     THE  MAXIMUM  BENDING  MOMENT. 

The  fundamental  equation  (4),  namely    —  =  M,   is   true   for 

any  section  of  any  beam,  /  being  the  moment  of  inertia  of  that 
section  about  its  neutral  axis,  c  the  vertical  distance  from  that 
axis  to  the  remotest  fiber,  S  the  tensile  or  compressive  unit-stress 
on  that  fiber,  and  M  the  bending  moment  of  all  the  external 
forces  on  one  side  of  the  section.  For  a  beam  of  constant  cross- 
section  5"  varies  directly  as  M,  and  the  greatest  S  will  be  found 
where  M  is  a  maximum.  The  place  where  M  has  its  maximum 
value  may  hence  be  called  the  '  dangerous  section,'  it  being  the 
section  where  the  horizontal  fibers  are  most  highly  strained. 

For  a  simple  beam  uniformly  loaded  with  w  pounds  per  linear 
unit  the  dangerous  section  is  evidently  at  the  middle,  and  as  shown 

wl2 

in  Art.  17,  the  maximum  M  is . 

8 

For  a  simple  beam  loaded  with  a  single  weight  Pat  the  distance 
/  from  the  left  support,  the  left  reaction  is  R  =  P — ~,  and  the 

maximum  moment  is  — I        P)P        If  p  be  movable  the  distance 


46  ON   SIMPLE   BEAMS   AND   CANTILEVERS.  III. 

p  will  be  variable  and  when  the  load  is  at  the  middle  the  maxi- 
mum M  is 

4' 
For  a  cantilever  the  dangerous  section  is  evidently  at  the  wall 

and  for  a  uniform  load  the  maximum  M  is  — . 

2 

For  a  beam  loaded  with  given  weights,  either  uniform  or  con- 
centrated, it  may  be  shown  that  the  dangerous  section  is  at  the 
point  where  the  vertical  shear  passes  through  zero.  To  prove 
this  let  P  be  any  concentrated  load  and  p  its  distance  from  the 
left  support,  and  w  the  uniform  load  per  linear  unit.  Then,  for 
any  section  distant  x  from  the  left  support, 

M=Rx—wx.  x_—IP(x—p\ 

To  find  the  value  of  x  which  renders  this  a  maximum  the  first 
derivative  must  be  put  equal  to  zero ;  thus, 

^=R-WX-IP=Q. 
dx 

But  R  —  wx —  2P  is  the  vertical  shear  Ffor  the  section  x  (see 
Art.  16).  Therefore  the  maximum  moment  occurs  at  the  section 
where  the  vertical  shear  passes  through  o.  To  find  the  dangerous 
section  an  expression  may  be  written  for  V  in  terms  of  x  and  the 
value  of  x  determined  when  V  changes  sign.  Thus  for  a  simple 
beam  uniformly  loaded  V  =•  %wl — wx  —  o,  and  x  =  l/2l.  For 
concentrated  loads  it  will  generally  be  necessary  to  find  by  trial 
the  point  where  the  shear  becomes  zero. 

Prob.  41.  A  simple  beam  12  feet  long  weighs  20  pounds  per 
foot  and  carries  a  load  of  100  pounds  at  4  feet  from  the  left  end 
and  a  load  of  50  pounds  at  7  feet  from  the  left  end.  Find  the 
dangerous  section. 

Prob.  42.  A  beam  25  feet  long,  uniformly  loaded  with  w  pounds 
per  linear  foot,  is  supported  at  the  left  end  and  at  a  point  5  feet 
from  the  right  end.  Find  the  two  dangerous  sections  and  the 
two  maximum  moments. 


ART.  24.  THE  INVESTIGATION  OF  BEAMS.  47 

ART.  24.-   THE  INVESTIGATION  OF  BEAMS. 

The  investigation  of  a  beam  consists  in  deducing  the  greatest 
horizontal  unit-stress  S  in  the  beam  from  the  fundamental  formula 
(4).  This  may  be  written, 

s  =  Mc 

First,  from  the  given  dimensions  find,  by  Art.  21,  the  value  of  c 
and  by  Art.  22  the  value  of  7.  Then  by  Art.  23  determine  the 
value  of  maximum  M.  From  (4)  the  value  of  ^  is  now  known. 
Usually  c  and  /  are  taken  in  inches,  and  M  in  pound-inches  ;  then 
the  value  of  6"  will  be  in  pounds  per  square  inch. 

The  value  of  5  will  be  tension  or  compression  according  as 
the  remotest  fiber  lies  on  the  concave  or  convex  side  of  the  beam. 
If  S'  be  the  unit-stress  on  the  opposite  side  of  the  beam  and 
c'  the  distance  from  the  neutral  axis,  then  from  law  (G\ 

?=?_  and  S'=S  — 

c         c'  c 

If  5  be  tension,  S1  will  be  compression,  and  vice  versa.  Some- 
times it  is  necessary  to  compute  S'  as  well  as  S  in  order  to 
thoroughly  investigate  the  stability  of  the  beam.  By  comparing 
the  values  of  S  and  S'  with  the  proper  working  unit-stresses  for 
the  given  materials  (see  Art.  8)  the  degree  of  security  of  the  beam 
may  be  inferred. 

As  an  example  consider  a  wrought  iron  I  beam  whose  depth 
is  12  inches,  width  of  flange  4.5  inches,  thickness  of  flange  I  inch 
and  thickness  of  web  0.78  inches.  It  is  supported  at  its  ends 
forming  a  span  of  12  feet,  and  carries  two  loads  each  of  10  qoo 
pounds  one  at  the  middle  and  the  other  at  one  foot  from  the 
right  end. 

By  Art.  14,  ^=6193  pounds. 

By  Art.  21,  c  =  6  inches. 

By  Art.  22,  7  —  338  inches4. 

By  Art.  23,  x  =  6  feet  for  dangerous  section. 

By  Art.  23,      max.  M=  36  078  X  12  pound-inches. 


48  ON   SIMPLE    BEAMS   AND    CANTILEVERS.  III. 

Then  from  formula  (4)  the  unit-stress  at  the  dangerous  section  is, 
5  =  36  078  X_I2_X_6  =  7  7QO  pounds  per  square  inch 

338 

This  is  the  compressive  unit-stress  on  the  upper  fiber  and  also 
the  tensile  unit-stress  on  the  lower  fiber,  and  being  only  about 
one-third  of  the  elastic  limit  for  wrought  iron  and  about  one- 
seventh  of  the  ultimate  strength  it  appears  that  the  beam  is  en- 
tirely safe.  It  will  usually  be  best  in  solving  problems  to  insert 
all  the  numerical  values  at  first  in  the  formula  and  thus  obtain 
the  benefit  of  cancellation. 

A  short  beam  heavily  loaded  also  needs  to  be  investigated  for 
the  shearing  stress  at  the  supports  in  the  manner  mentioned  in 
Art.  20,  but  in  ordinary  cases  there  is  little  danger  of  failure 
from  this  cause. 

Prob.  43.  Find  the  factor  of  safety  of  a  simple  wooden  beam 
2X4  inches  when  loaded  at  the  middle  with  i  ooo  pounds. 

Prob.  44.  A  piece  of  scantling  2  inches  square  and  10  feet  long 
is  hung  horizontally  by  a  rope  at  each  end  and  three  painters 
stand  upon  it.  Is  it  safe  ? 

Prob.  45.  A  wrought  iron  bar  one  inch  in  diameter  and  two 
feet  long  is  supported  at  its  middle  and  a  load  of  500  pounds 
hung  upon  each  end  of  it.  Find  its-  factor  of  safety. 

ART.  25.     SAFE  LOADS  FOR  BEAMS. 

The  proper  load  for  a  beam  should  not  make  the  value  of  6"  at 
the  dangerous  section  greater  than  the  allowable  unit-stress.  This 
allowable  unit-stress  or  working  strength  is  to  be  assumed  accord- 
ing to  the  circumstances  of  the  case  by  first  selecting  a  suitable 
factor  of  safety  from  Art.  8  and  dividing  the  ultimate  strength  of 
the  material  by  it,  the  least  ultimate  strength  whether  tensile  or 
compressive  being  taken.  For  any  given  beam  the  quantities  / 
and  c  are  known.  Then,  by  the  general  formula  (4),  the  bending 
moment  M  may  be  expressed  in  terms  of  the  unknown  loads  on 


108 

ART.    26.  DESIGNING    OF    BEAMS.  49 

the  beam,  and  thus  those  loads  be  found.  The  sign  of  the  bend- 
ing moment  should  not  be  used  in  (4),  since  that  sign  merely  de- 
notes whether  the  upper  fiber  of  the  beam  is  in  tension  or  com- 
pression, or  indicates  the  direction  in  which  the  external  forces 
tend  to  bend  it. 

As  an  example,  consider  a  cantilever  projecting  from  a  wall 
whose  length  is  6  feet,  breadth  2  inches,  depth  3  inches  and  which 
is  loaded  uniformly  with  w  pounds  per  linear  foot.  It  is  required 
to  find  the  value  of  w  so  that  vS  may  be  800  pounds  per  square 
inch.  Here  we  have  c  =  i  finches,  /=§,  and  M=  36  X  6w. 
Then  from  (4), 

800  X  54 

210  zu  = — -, — ; '          whence         w=  n  pounds. 

I#  X   12 

Since  a  wooden  beam  2X3  inches  weighs  about  2  pounds  per 
linear  foot,  the  safe  load  in  this  case  will  be  about  9  pounds 
per  foot. 

Prob.  46.  A  wooden  beam  8X9  inches  and  of  14  feet  span 
carries  a  load,  including  its  own  weight,  of  w  pounds  per  linear 
foot.  Find  the  value  of  w  for  a  factor  of  safety  of  10. 

Prob.  47.  A  cast  iron  beam  one  inch  square  and  of  4  feet  span 
carries  a  load  P  at  the  middle.  Find  P  so  that  the  greatest  hori- 
zontal unit-stress  at  the  dangerous  section  shall  be  3  ooo  pounds 
per  square  inch. 

ART.  26.      DESIGNING  OF  BEAMS. 

When  a  beam  is  to  be  designed  the  loads  to  which  it  is  to  be 
subjected  are  known,  as  also  is  its  length.  Thus  the  maximum 
bending  moment  may  be  found.  The  allowable  working  strength 
5  is  assumed  in  accordance  with  engineering  practice.  Then 
formula  (4)  may  be  written, 

c         S 
and  the  numerical  value  of  the  second  member  be  found.     The 


5O  ON   SIMPLE    BEAMS    AND    CANTILEVERS.  III. 

dimensions  to  be  chosen  for  the  beam  must  then  have  a  value  of 
—  equal  to  this  numerical  value,  and  these  in  general  are  deter- 
mined tentatively,  certain  proportions  being  first  assumed.  The 
selection  of  the  proper  proportions  and  shapes  of  beams  for  differ- 
ent cases  requires  much  judgment  and  experience.  But  what- 
ever forms  be  selected  they  must  in  each  case  be  such  as  to  satisfy 
the  above  equation. 

For  instance,  a  wrought  iron  beam  of  4  feet  span  is  required  to 
carry  500  pounds  at  the  middle.  Here,  by  Art.  23,  the  value  of 
maximum  Mis  6  ooo  pound-inches.  From  Art.  8  the  value  of  5" 
for  a  variable  load  is  about  10  ooo  pounds  per  square  inch.  Then, 

7=    6000    =o6inches3. 

c         10  ooo 
An  infinite  number  of  cross-sections  may  be  selected  with  this 

value  of  — .    If  the  beam  is  to  be  round  and  of  diameter  d,  it  is 

known  that  c.  =  l/2d  and    7=^.     Hence, 

64 

—  =0.6,  whence  d=  1.83  inches. 

32 
If  the  cross-section  is  to  be  rectangular,  the  dimensions  I  X  2 

inches  would  give  the  value  of  —    as  ^3  which  would  be  a  little 

too  large,  while  I  X  i^  would  give  —  as  about  0.53  which  would 
be  too  small. 

Sometimes  cross-sections  not  symmetrical  to  the  neutral  axis  are 
designed,  particularly  for  cast  iron  where  the  compressive  unit- 
stress  may  be  taken  greater  than  the  tensile.  In  no  case  of  design 
however  should  the  dimensions  be  so  selected  as  to  render  the 
unit-stress  on  either  side  greater  than  the  elastic  limit  of  the 
material. 

Prob.  48.  A  rectangular  beam  of  14  feet  span  carries  a  load  of 
i  ooo  pounds  at  its  center.  If  its  width  is  4  inches  find  its  depth 
for  a  factor  of  safety  of  10. 


ART.  27. 


THE    MODULUS   OF    RUPTURE. 


Prob.  49.  Design  a  hollow  circular  wrought  iron  beam  for  a 
span  of  12  feet  to  carry  a  load  of  320  pounds  per  linear  foot. 

ART.  27.     THE  MODULUS  OF  RUPTURE. 

The  fundamental  formula  (4)  is  only  true  for  stresses  within  the 
elastic  limit,  since  beyond  that  limit  the  law  (G)  does  not  hold, 
and  the  horizontal  unit-stresses  are  no 
longer  proportional  to  their  distances 
from  the  neutral  axis,  but  increase  in  a 
more  rapid  ratio  as  shown  in  the 
sketch.  It  is  however  very  customary 
to  apply  (4)  to  the  rupture  of  beams.  Figiir. 

The  'modulus  of  rupture '  is  the  value  of  S  deduced  from  for- 
mula (4)  when  the  beam  is  loaded  up  to  the  breaking  point.  It  is 
always  found  by  experiment  that  the  modulus  of  rupture  does 
not  agree  with  either  the  ultimate  tensile  or  compressive  strength 
of  the  material  but  is  intermediate  between  them.  If  formula  (4) 
were  valid  beyond  the  elastic  limit,  the  value  of  .S  for  rupture 
would  agree  with  the  least  ultimate  strength,  with  tension  in  the 
case  of  cast  iron  and  with  compression  in  the  case  of  timber.  The 
modulus  of  rupture  is  denoted  by  Sr. 

The  average  values  of  the  modulus  rupture  are  given  in  the 
following  table,  which  also  contains  the  average  ultimate  tensile 
and  compressive  strengths,  previously  stated  in  Arts.  5  and  6,  all 
in  pounds  per  square  inch. 


Material. 

Tensile 
Strength,  St. 

Modulus  of 
Rupture,  Sr. 

Compressive 
Strength,  8C. 

Timber 

IO  OOO 

9  ooo 

8  000 

Brick 

2    500 

Stone 

2   OOO 

6  ooo 

Cast  Iron 

2O  OOO 

35  ooo 

90  ooo 

Wrought  Iron 

55  ooo 

55  ooo 

55  ooo 

Steel 

IOO  OOO 

120  000 

150  ooo 

52  ON   SIMPLE    BEAMS   AND    CANTILEVERS.  III. 

By  the  use  of  the  experimental  values  of  the  modulus  of  rup- 
ture it  is  easy  with  the  help  of  formula  (4)  to  determine  what  load 
will  cause  the  rupture  of  a  given  beam,  or  what  must  be  its  length 
or  size  in  order  that  it  may  rupture  under  assigned  loads. 

Prob.  50.  A  wooden  beam  4X6  inches  and  9  feet  span  has  a 
load  P  at  the  middle.  Find  the  value  of  P  to  break  it. 

Prob.  51.  A  wrought  iron  cantilever  2  inches  square  projects 
from  a  wall.  Find  its  length  in  order  to  rupture  under  its  own 
weight. 

Prob.  52.  What  must  be  the  size  of  a  square  wooden  beam  of 
8  feet  span  in  order  to  break  under  its  own  weight  ? 


ART.  28.     COMPARATIVE  STRENGTHS. 

The  strength  of  a  beam  is  measured  by  the  load  that  it  can 
carry.  Let  it  be  required  to  determine  the  relative  strength  of 
the  four  following  cases. 

1st,  A  cantilever  loaded  at  the  end  with  W, 

2nd,  A  cantilever  uniformly  loaded  with  W, 

3rd,  A  simple  beam  loaded  at  the  middle  with  W, 

4th,  A  simple  beam  loaded  uniformly  with  W. 

Let  /  be  the  length  in   each   case.     Then,  from  Art.  23   and 

formula  (4), 

SI 
Ic  ' 
SI 
l~k' 
SI 

I7 

Therefore  the  comparative  strengths  of  the  four  cases  are  as  the 
numbers  i,  2,  4,  8.  That  is,  if  four  such  beams  be  of  equal  size 
and  length  and  of  the  same  material,  the  2nd  is  twice  as  strong 


For  ist, 
For  2nd, 
For  3rd, 
For  4th, 

M=  Wl 

**? 

M=™L 

4 

Wl 

Q 

and  hence 
and  hence 
and  hence 
and  hence 

ART.    29.  RECTANGULAR    BEAMS.  53 

as  the  ist,  the  3rd  four  times  as  strong,  and  the  4th  eight  times 
as  strong. 

These  equations  also  show  that  the  strength  of  a  beam  varies 
directly  as  S,  inversely  as  the  length  /,  and  directly  as  -.  They 

also  prove  that  a  uniform  load  produces  only  one-half  as  much 
stress  as  an  equivalent  concentrated  load.  These  general  laws  of 
the  strength  of  beams  are  very  important. 

Prob.  53.  Compare  the  strength  of  a  rectangular  beam  2  inches 
wide  and  4  inches  deep  with  that  of  a  circular  beam  3  inches  in 
diameter. 

Prob.  54.  Compare  the  strength  of  a  wooden  beam  4X6  inches 
and  10  feet  span  with  that  of  a  wrought  iron  beam  I  X  2  inches 
and  7  feet  span. 

ART.  29.     RECTANGULAR  BEAMS. 

For  a  beam  of  rectangular  cross-section  let  b  be  the  breadth 
and  d  the  depth.  Then  the  formula  (4)  becomes, 


For  cantilevers  and  common  beams  let  W  be  the  load,  either 
uniform  or  concentrated  as  in  Art.  28,  then, 
...        Sbd* 

w=n-6T' 

where  n  is  either  I,  2,  4  or  8,  as  the  case  may  be. 

This  equation  shows  the  important  laws  that  the  strength  of  a 
rectangular  beam  varies  directly  as  its  breadth,  directly  as  the 
square  of  its  depth  and  inversely  as  its  length.  The  reason  xwhy 
rectangular  beams  are  put  with  the  greatest  dimension  vertical  is 
now  apparent. 

To  find  the  strongest  rectangular  beam  that  can  be  cut  from  a 
circular  log  of  given  diameter  D,  it  is  necessary  to  make  bd*  a 
maximum.  Or  the  value  of  b  is  to  be  found  which  makes  b(D*  —  £*) 


54 


ON    SIMPLE    BEAMS    AND    CANTILEVERS. 


III. 


a  maximum.     By  placing  the  first  derivative  equal  to  zero  this 

value  of  b  is  readily  found.     Thus, 

b  =  D\/y-,  and  d=D\/%. 

Hence  very  nearly,  b:  d  :  :  5  /  7.  From  this  it  is  evident  that  the 
way  to  lay  off  the  strongest  beam  on  the  end  of 
a  circular  log  is  to  divide  the  diameter  into  three 
equal  parts,  from  the  points  of  division  draw 
perpendiculars  to  the  circumference,  and  then  join 
the  points  of  intersection  with  the  ends  of  the 
ng.a  diameter,  as  shown  in  the  figure. 

Prob.  55.  Compare  the  strength  of  a  cylindrical  beam  with  that 
of  the  strongest  rectangular  beam  that  can  be  cut  from  it. 

ART.    30.      WROUGHT  IRON  I  BEAMS. 

Wrought  iron  I  beams  are  rolled  at  present  in  about  thirteen 
sizes  or  different  depths.  Of  each  size  there 
is  a  light  and  a  heavy  weight,  and  by  giving 
special  orders  weights  intermediate  in  value 
may  be  obtained.  They  are  extensively  used 
in  engineering  and  architecture.  The  follow- 
ing table  gives  the  sizes,  weights  and  moments 
of  inertia  of  those  manufactured  by  Carnegie 
Bros.  &  Co.,  Pittsburgh,  Pa.  The  sizes  of 
different  manufacturers  agree  as  to  depth,  but 
vary  slightly  with  regard  to  proportions  of 
cross-section,  weights  per  foot,  and  moments 

of  inertia.       Fig.    19  shows  the  proportions  of  the  light  and 

heavy  6  inch  beams. 

The  moments  of  inertia  in  the  fourth  column  of  the  table  are 
taken  about  an  axis  perpendicular  to  the  web  at  the  center,  this 
being  the  neutral  axis  of  the  cross-section  when  used  as  a  beam. 
The  values  of  /'  are  for.  use  in  Chapter  V.  In  investigating  the 

strength  of  a  given  I  beam  the  value  of-  is  taken  from  the  table 

c 


Tig.  19. 


ART.  30. 


WROUGHT    IRON    I    BEAMS. 


55 


and  5  is  computed  from  formula  (4).     In  designing  an  I  beam 
for  a  given  span  and  loads  the  value  of  -   is  found  by  (4)  from 

the  data  and  then  from  the  table  that  I  is  selected  which  has  the 
nearest  or  next  highest  corresponding  value. 


Size. 
Depth. 
Inches. 

Width  of 
Flange. 
Inches. 

Weight 
per  foot. 
Pounds. 

/. 

Inches*. 

/ 
c 
Inches3. 

/'. 

Inches4. 

Heavy  15 

5.81 

80 

750 

IOO. 

29.9 

Light     1  5 

5-55 

67 

677 

9°-3 

25.4 

H          I5 

5-33 

65 

614 

81.9 

2O.O 

L           15 

5-03 

50 

53° 

70.6 

16-3 

H              12 

5.09 

60 

340 

56.7 

15-5 

L               12 

4.64 

42 

275 

45-9 

II.O 

H          io# 

4.92 

45 

201 

38.3 

10.7 

L           10% 

4-54 

3»# 

I65 

3i-4 

8.01 

H          10 

4-77 

45 

I87 

37-5 

"•3 

L          10 

4.32 

3° 

150 

30.0 

7-94 

H           9 

4-33 

33 

117 

26.0 

7.14 

L            9 

4.01 

23^ 

97-5 

21.7 

5.48 

H           8 

4.29 

35 

90.4 

22.6 

6.96 

L            8 

3-8i 

22 

69.6 

17-5 

4-57 

H           7 

3-91 

25 

54-3 

15-5 

4.87 

L            7 

3-6i 

18 

45.8 

I3-I 

3-72 

H           6 

3-46 

18 

28.4 

9.48 

2.51 

L            6 

3-24 

i*# 

24.5 

8.16 

2.00 

H            5 

2.91 

13 

14.2 

5.69 

i-34 

L            5 

2-73 

IO 

12.3 

4-94 

i.  08 

H           4 

2.63 

10 

6.99 

3-50 

0.87 

L            4 

2.48 

8 

6.19 

3.10 

0.71 

H           3 

2.52 

9 

3-54 

2.36 

0.84 

L            3 

2.32 

7 

3-09 

2.06 

0.55 

For  example,  let  it  be  required  to  determine  which  I  should 
be  selected  for  a  floor  loaded  with  150  pounds  per  square  foot, 
the  beams  to  be  of  20  feet  span  and  spaced  1 2  feet  apart  between 
centers,  and  the  maximum  unit-stress  S  to  be  12  ooo  pounds  per 
square  inch.  Here  the  total  uniform  load  on  the  beam  is 


ON    SIMPLE    BEAMS   AND    CANTILEVERS. 


III. 


12  X  2O  X  150=  36  ooo  pounds  =  W.     From  formula  (4), 

I _M  _  36  ooo  X  20  X  12 
c       S 


=  90. 


8X  12  ooo 
and  hence  from  the  table,  the  light  15  inch  I  should  be  selected. 

Steel  I  beams  and  the  other  shapes  are  now  beginning  to  be 
used,  and  will  undoubtedly  be  very  common  in  a  few  years. 

Prob.  56.  A  heavy  15  inch  I  beam  of  12  feet  span  sustains  a 
uniformly  distributed  load  of  41  net  tons.  Find  its  factor  of  safety. 
Also  the  factor  of  safety  for  a  24  feet  span  under  the  same  load. 

Prob.  57.  A  floor,  which  is  to  sustain  a  uniform  load  of  175 
pounds  per  square  foot,  is  to  be  supported  by  heavy  10  inch  I 
beams  of  1 5  feet  span.  Find  their  proper  distance  apart  from 
center  to  center  so  that  the  maximum  fiber  stress  may  be  1 2  ooo 
pounds  per  square  inch. 

Prob.  58.  What  I  should  be  selected  for  this  floor  if  the  beams 
are  to  be  spaced  3  feet  7  inches  from  center  to  center. 

ART.  31.     WROUGHT  IRON  DECK  BEAMS. 

Deck  beams  are  used  in  the  construction  of  buildings,  and  are 
of  a  section  such  as  shown  in  figure   20. 

O° 
The  heads  are  formed  with  arcs  of  circles 
but  may  be  taken  as  elliptical  in  computing 
the  values  of  c  and  /.  The  following  table 
gives  dimensions  of  deck  beams  manufac- 
tured by  Carnegie  Bros.  &  Co. 

By  means  of  formula  (4)  a  given  deck 
beam  may  be  investigated  or  safe  loads  be 
determined  for  it,  or  one  may  be  selected  for 
a  given  load  and  span.  Sometimes  T  irons 
are  used  instead  of  deck  beams  ;  the  values 
of  c  and  /  for  these  may  be  computed  with 
sufficient  accuracy  by  regarding  the  web  and  flanges  as  rect- 
angular as  in  Arts.  21  and  22. 


ART.  32. 


CAST    IRON    BEAMS. 


57 


Size.  Depth. 

Width 
of 

Thickness 

Weight 

c. 

/. 

/ 

Inches. 

Flange. 

of  Web. 

per  ft. 

c. 

Inches. 

Inches. 

Fnds. 

Inches. 

Inch's*. 

Inch's*. 

Heavy   9 

3-97 

0.625 

30 

4-59 

9I.9 

20.O 

Light     9 

3-75 

0.406 

23^ 

4.60 

78.6 

I7.I 

H            8 

4.00 

0.750 

28 

449 

63.3 

I4.I 

L            8 

3-75 

0.500 

21% 

4.58 

52.1 

1  1.6 

H           7 

3-75 

0.625 

23 

3-98 

43-0 

10.8 

L            7 

3-50 

0-375 

17 

4.00 

34-4 

8.6 

Prob.  59.  A  heavy  7"  deck  beam  is  loaded  uniformly  with 
50  ooo  pounds.     Find  its  factor  of  safety  for  a  span  of  22  feet 


ART.  32.     CAST  IRON  BEAMS. 

Wrought  iron  beams  are  usually  made  with  equal  flanges  since 
the  resistance  of  wrought  iron  is  about  the  same  for  both  tension 
and  compression.  For  cast  iron,  however,  the  flange  under  ten- 
sion should  be  larger  than  that  under  compression,  since  the 
tensile  resistance  of  the  material  is  much  less  than  its  compressive 
resistance.  Let  S'  be  the  unit-stress  on  the  remotest  fiber  on  the 
tensile  side  and  5  that  on  the  compressive  side,  at  the  distances 
c'  and  c  respectively  from  the  neutral  axis.  Then,  from  law  (G), 

c-  =? 
c'       S1' 

Now  if  the  working  values  of  5"  and  S'  can  be  selected  the  ratio 
of  c  to  c'  is  known  and  a  cross-section  can  be  designed,  but  it  is 
difficult  to  assign  these  proper  values  on  account  of  our  lack  of 
knowledge  regarding  the  elastic  limits  of  cast  iron. 

According  to  Hodgkinson's  investigations  the  following  are 
dimensions  for  a  cast  iron  beam  of  equal  ultimate  strength. 
Thickness  of  web  =         t, 

Depth  of  beam  =  13.5*, 


58  ON    SIMPLE    BEAMS    AND    CANTILEVERS.  III. 

Width  of  tensile  flange  =     I2/, 

Thickness  of  tensile  flange  =       2t, 

Width  of  compressive  flange  =       $t, 

Thickness  of  compressive  flange  =  i%t, 

Value  of  c  —       gf, 

Value  of  7  =  923/4. 

Here  the  unit-stress  in  the  tensile  flange  is  one-half  that  in  the 
the  compressive  flange.  Although  these  proportions  may  be 
such  as  to  allow  the  simultaneous  rupture  of  the  flanges,  yet  it 
does  not  necessarily  follow  that  they  are  the  best  proportions  for 
ordinary  working  stresses,  since  the  factors  of  safety  in  the  flanges 
as  computed  by  the  use  of  formula  (4)  would  be  quite  different. 
The  proper  relative  proportions  of  the  flanges  of  cast  iron  beams 
for  safe  working  stresses  have  never  been  definitely  established, 
and  on  account  of  the  extensive  use  of  wrought  iron  the  question 
is  not  now  so  important  as  formerly. 

As  an  illustration  of  the  application  of  formula  (4)  let  it  be  re- 
quired to  determine  the  total  uniform  load  W  for  a  cast  iron  1 
beam  of  14  feet  span,  so  that  the  factor  of  safety  may  be  6,  the 
depth  of  the  beam  being  1 8  inches,  the  width  of  the  flange  1 2 
inches,  the  thickness  of  the  stem  i  inch,  and  the  thickness  of  the 
flange  i^  inches.  First,  from  Art.  21  the  value  of  c  is  found  to 
be  12.63  inches,  and  that  of  c'  to  be  5.37  inches.  From  Art.  22 
the  value  of  /  is  computed  to  be  I  031  inches4.  From  Art.  23 
the  bending  moment  is, 

M  =  ——  =  21  W7  pound-inches. 
8 

Now  with  a  factor  of  safety  of  6  the  working  strength  5  on  the 
the  remotest  fiber  of  the  stem  at  the  dangerous  section  is  to  be 

2 pounds  per  square  inch.     Hence  from  formula  (4), 

6 

21  W  =  90QQQX  1031,  whence    w  = 

6  X  12.63 
Again  with  a  factor  of  safety  of  6  the  working  strength  5'  on 


ART.    33.       GENERAL  EQUATION  OF  THE  ELASTIC  CURVE.  59 

the  remotest  fiber  of  the  flange  at  the  dangerous  section  is  to  be 

pounds  per  square  inch.     Hence  from  the  formula, 

6 

2 1  W  =  20QO°X  l°l\    whence    W=  30 400 pounds. 

The  total  uniform  load  on  the  beam  should  hence  not  exceed 
30  400  pounds.  Under  this  load  the  factor  of  safety  on  the  tensile 
side  is  6,  while  on  the  compressive  side  it  is  nearly  12. 

Prob.  60.  A  cast  iron  beam  in  the  form  of  a  channel,  or  hollow 
half  rectangle,  is  often  used  in  buildings.  Suppose  the  thickness 
to  be  uniformly  one  inch,  the  base  8  inches,  the  height  6  inches 
and  the  span  1 2  feet.  Find  the  values  of  5  and  S'  at  the  danger- 
ous section  under  a  uniform  load  of  1 5  net  tons. 

ART.  33.     GENERAL  EQUATION  OF  THE  ELASTIC  CURVE. 

When  a  beam  bends  under  the  action  of  exterior  forces  the 
curve  assumed  by  its  neutral  surface  is  called  the  elastic  curve. 
It  is  required  to  deduce  a  general  expression  for  its  equation. 

Let  pu  in  the  figure  be  any  normal  section  in  any  beam.  Let 
mn  be  any  short  length  dl, 
and  draw  vmq  parallel  to  the 
normal  section  through  n. 
Previous  to  the  bending  the 
sections  pu  and  st  were  par- 
allel ;  now  they  intersect  at  o 
the  center  of  curvature.  Pre- 
vious to  the  bending  ps  was 
equal  to  dl,  now  it  has 

elongated  the  amount  pq  or  Fig2i 

L     The  distance  mp  is  the 

quantity  c.  The  elongation  ),  is  produced  by  the  unit-stress  5, 
and  from  (2)  its  value  is, 

<-¥ 


60  ON    SIMPLE    BEAMS   AND    CANTILEVERS.  III. 

where  E  is  the  coefficient  of  elasticity  of  the  material  of  the  beam. 
From  the  similar  figures  omn  and  mpq, 

om  =  mp  ^  or  *  =  -, 

mn        pq  '  dl        £ 

where  R  is  the  radius  of  curvature  om.  Inserting  in  this  the 
above  value  of  /,  it  becomes, 

S  _E 
~c~  R' 
But  from  the  fundamental  formula  (4), 

5        M 
c  ~:    /' 
and  hence,  by  comparison, 

El 

M  =  _.. 

This  is  the  formula  which  gives  the  relation  between  the  bending 
moment  of  the  exterior  forces  and  the  radius  of  curvature  at 
any  section. 

Now,  in  works  on  the  differential  calculus,  the  following  value 
is  deduced  for  the  radius  of  curvature  of  any  plane  curve  whose 
abscissa  is  ;r,  ordinate  y  and  length  /,  namely, 


dx.d*y 
Hence  the  most  general  equation  of  the  elastic  curve  is, 

dl*        =  £7 
dx.d*y    =~~  M' 

which  applies  to  the  flexure  of  all  bodies  governed  by  the  laws 
of  Arts.  3  and  18. 

In  discussing  a  beam  the  axis  of  x  is  taken  as  horizontal  and 
that  of  y  as  vertical.  Experience  teaches  us  that  the  length  of  a 
small  part  of  a  bent  beam  does  not  materially  differ  from  that  of 
its  horizontal  projection.  Hence  dl  may  be  placed  equal  to  dx 
for  all  beams,  and  the  above  equation  reduces  to  the  form, 
d*y  M  (  } 

~  ~     ~      ' 


ART.    34.  DEFLECTION   OF   CANTILEVERS.  6 1 

This  is  the  general  equation  of  the  elastic  curve,  applicable  to 
all  beams  whatever  be  their  shapes,  loads  or  number  of  spans. 
M  is  the  bending  moment  of  the  external  forces  for  any  section 
whose  abscissa  is  x,  and  whose  moment  of  inertia  with  respect 
to  the  neutral  axis  is  /.  Unless  otherwise  stated  /will  be  regarded 
as  constant,  that  is,  the  cross-section  of  the  beam  is  constant 
throughout  its  length. 

To  obtain  the  particular  equation  of  the  elastic  curve  for  any 
special  case,  it  is  first  necessary  to  express  M  as  a  function  of  x 
and  then  integrate  the  general  equation  twice.  The  ordinate  y 
will  then  be  known  for  any  value  of  x.  It  should  however  be 
borne  in  mind  that  formula  (5),  like  formula  (4),  is  only  true 
when  the  unit-stress  S  is  less  than  the  elastic  limit  of  the  material. 

Prob.  61.  A  wooden  beam  y2  inch  wide,  ^  inch  deep,  and  3 
feet  span  carries  a  load  of  14  pounds  at  the  middle.  Find  the 
radius  of  curvature  for  the  middle,  quarter  points  and  ends. 

ART.  34.     DEFLECTION  OF  CANTILEVERS. 

Case  I.  A  load  at  the  free  end. — Take  the  origin  of  co-ordi- 
nates at  the  free  end,  and  as 
in  Fig.  22,  let  m  be  any  point 
of  the  elastic  curve  whose 
abscissa  is  x  and  ordinate  y. 
For  this  point  the  bending 
moment  M  is  —  Px  and  the  general  formula  (5)  becomes, 


By  integration, 


dy  Px* 

EIdx  =  --T~ 


But  —  is  the  tangent  of  the  angle  which  the  tangent  to  the 

dx 
elastic  curve  at  m  makes  with  the  axis  of  x  and  as  the  beam  is 

fixed  at  the  wall  the  value  of   —   is  o  when  x  equals  /.      Hence 

dx 


62  ON   SIMPLE    BEAMS   AND    CANTILEVERS.  III. 


C  =  y^Pl2,  and  the  first  differential  equation  is, 
EJdy  __    PI2   __  Px* 

dx  2  2 

The  second  integration  now  gives, 

Pl*x       Px* 
Ely  =—---+?. 

But  y  =  o,  when  x  =  o.     Hence  O  =  o,  and 

6Efy  =  P  (3/2*  —  *3), 

which  is  the  equation  of  the  elastic  curve  for  a  cantilever  of  length 
/  with  a  load  P  at  the  free  end.  If  x  =  I  the  value  of  y  will  be 
the  maximum  deflection,  which  may  be  represented  by  J.  Then, 


and  for  any  point  of  the  beam  the  deflection  is  J  —  y. 

Case  II.  A   cantilever     uni- 
formly loaded. — Let  the  origin 
be  taken  at  the  free  end  as  be- 
fore.    Let  the  load    per  linear 
rig>-M>  unit  be  w.     Then  for  any  sec- 

tion M  =  —  y^wx*  and  formula  (5)  becomes, 

~dx~2  =    ~2~' 

Integrate  this,  determine  the  constant  of  integration  by  the  con- 
sideration that  -^    =  o  when  x  =  /,  and  then, 
dx 


dx 

Intergrate  again,  and  after  determining  the  constant,  the  equation 
of  the  elastic  curve  is, 

which  is  a  biquadratic  parabola.     For  x  =  /,  y  =  J  the  maxi- 
mum deflection,  whose  value  is, 

"=£-w- 

where  W  is  the  total  uniform  load  on  the  cantilever. 


ART.    35.  DEFLECTION   OF   SIMPLE   BEAMS.  63 

Case  III.    A  cantilever  uniformly  loaded  with    W  and   also 

carrying  a  load  P  at  the  free  end.  —  Here  it  is  easy  to  show  that 

the  maximum  deflection  is, 

_ 

In  all  these  cases'  the  maximum  unit-stress  J>  produced  by  the 
loads  must  not  exceed  the  elastic  limit  of  the  material. 

Prob.  62.  Compute  the  deflection  of  a  wooden  cantilever, 
2X2  inches  and  6  feet  span,  caused  by  a  load  of  100  pounds  at 
the  end.  Also  of  a  cast  iron  cantilever  of  the  same  dimensions. 

Prob.  63.  In  order  to  find  the  coefficient  of  elasticity  of  a  cast 
iron  bar  2  inches  wide,  4  inches  deep  and  6  feet  long,  it  was  bal- 
anced upon  a  support  and  a  weight  of  4  ooo  pounds  hung  at 
each  end,  when  the  deflection  of  the  ends  was  observed  to  be 
0.401  inches.  Compute  the  value  of  E. 

ART.  35.     DEFLECTION  OF  SIMPLE  BEAMS. 

Case  I.  A  single  load  P  at  the  middle.  —  Let  the    origin  be 
taken  at   the   left   support. 
For  any   section    between          ^t—  -x— 
the   left  support    and    the 
middle  the  bending  moment 
M  is  y2Px.     Then  the  general  formula  (5)  becomes, 

EI&  =  *r. 

dx*  2 

Integrate  this  and  find  the  constant  by  the  fact  that  —  —  o 

dx 

when  x  =  y^.     Then  integrate  again  and  find  the  constant  by 
the  fact  that  y  =  o  when  x  —  o.     Thus, 


is  the  equation  of  elastic  curve  between  the  left  hand  support  and 
the  load.     For  the  greatest  deflection  make  x  =  }^l,  then, 


64  ON   SIMPLE    BEAMS    AND    CANTILEVERS.  III. 

This  result  may  also  be  obtained  by  regarding  the  beam  as  a  can- 
tilever fixed  at  the  load  and  bent  upward  by  the  reaction. 

Case  II.  A  uniform  load.  —  Let  w  be  the  load  per  linear  unit, 
then  the  formula  (5)  becomes, 

wlx        wx* 


Integrate  this  twice,  find  the  constants  as  in  the  preceding  para- 
graph, and  the  equation  of  the  elastic  curve  is, 

2^Efy  =  w(  —  x*  +  2/^3  —  frx), 
from  which  the  maximum  deflection  is, 


= 


384  384^7 

Case  III.  A  load  P  at  any  point.  —  Here  it  is  necessary  first  to 
consider  that  there  are  two  elastic  curves,  one  on  each  side  of  the 

load,   which   have    distinct 
&  ----------  »T  >^^£  equations,  but  which  have  a 

common  tangent  and  ordi- 
Fig:25-  nate  under  the  load.    As  in 

Fig.  25,  let  the  load  be  placed  at  a  distance  k  from  the  left  sup- 

port.    Then  the  left  reaction  is  P          .      From   the    general 
formula  (5),  with  the  origin  at  the  left  support,  the  equations  are, 

On  the  left  of  the  load, 

dzv 

(a)  £!•£  =  Rx, 

(b)  EI^  =  y2Rx*+  C^ 

(c)  Ely  =\Rx*    +£>+£. 
On  the  right  of  the  load, 

(a)'       EI=       Rx-P(x-k\ 


(c}r  Ely  =    \Rx*  —  j 


ART.    36.       COMPARATIVE   DEFLECTION   AND  STIFFNESS.  65 

To  determine  the  constants  consider  in  (c)  that  y  =  o  when 
x  =  o  and  hence  that  C3  =  o.  Also  in  (<:)',  y  =  o  when  x  •=  I 
and  hence  C4  is  known  in  te'rms  of./?,  Pand  C2.  Since  the  curves 
have  a  common  tangent  under  the  load,  (b)  =  (£>)'  when  x  =  k, 
and  since  they  have  a  common  ordinate  at  that  point  (c)  =  (c)f 
when  x  =  k.  Or, 

^Rk*  +  C^  =  ^Rk*  —  ^Pk*  +  Pfc     +  C9, 


From  these  two  equations  the  values  of  Ct  and   £72  are  found. 
Then  the  equation  of  the  elastic  curve  on  the  left  of  the  load  is, 


6EIy  =  Rx*  +  Pk($k  —  2l—  -)  x. 
To  find  the  maximum  deflection,  insert  in  this  the  value  of  R  and 

find  the  value  of  x  for  which  —  becomes   o.      If  k  be   greater 
dx 

than  y2l  this  value  of  x  inserted  in  the  above  equation  gives  the 
maximum  value  of  y.  If  k  be  less  than  y2l  the  maximum  deflec- 
tion is  on  the  other  side  of  the  load.  For  instance,  if  k  =  ^/, 
the  equation  of  the  elastic  curve  on  the  left  of  the  load  is, 


This  is  a  maximum  when  x  =  o.56/,  which  is  the  point  of  greatest 
deflection. 

Prob.  64.  In  order  to  find  the  coefficient  of  elasticity  of  Quercus 
alba  a  bar  4  centimeters  square  and  one  meter  long  was  supported 
at  the  ends  and  loaded  in  the  middle  with  weights  of  50  and  100 
kilograms  when  the  corresponding  deflections  were  found  to  be 
6.6  and  13.0  millimeters.  Show  that  the  mean  value  of  £  was 
74  500  kilos  per  square  centimeter. 

ART.  36.      COMPARATIVE  DEFLECTION  AND  STIFFNESS. 

From  the  two  preceding  articles  the  following  values  of  the 
maximum  deflections  may  now  be  written  and  their  comparison 
will  show  the  relative  stiffness  of  the  different  cases. 


66  ON    SIMPLE    BEAMS   AND   CANTILEVERS.  III. 

For  a  cantilever  loaded  at  the  end  with    W,        j  —  -  --- 

For  a  cantilever  uniformly  loaded  with   W,         A  =  -.  -------  • 

Wfi 
For  a  simple  beam  loaded  at  middle  with  W,        J  =  —  .  __ 

For  a  simple  beam  uniformly  loaded  with  W,         J  =  -5_.  __ 

The  relative  deflections  of  these  four  cases  are  hence  as  the  num- 
bers i,|.  ^  and^. 

These  equations  also  show  that  the  deflections  vary  directly 
as  the  load,  directly  as  the  cube  of  the  length  and  inversely  as 

E  and  /.     For  a  rectangular  beam  /  =   —  ,  and  hence  the  de- 

12 

flection  of  a  rectangular  beam  is  inversely  as  its  breadth  and  in- 
versely as  the  cube  of  its  depth. 

The  stiffness  of  a  beam  is  indicated  by  the  load  that  it  can 
carry  with  a  given  deflection.  From  the  above  it  is  seen  that  the 
value  of  the  load  is, 


where  m  has  the  value  3,  8,  48,  or  —  as  the  case  may  be.  There- 
fore, the  stiffness  of  a  beam  varies  directly  as  E,  directly  as  /  and 
inversely  as  the  cube  of  its  length,  and  the  relative  stiffness  of 
the  above  four  cases  is  as  the  numbers  I,  2->  16  and  25-.  From 
this  it  appears  that  the  laws  of  stiffness  are  very  different  from 
those  of  strength.  (Art.  28.) 

Prob.  65.  Compare  the  strength  and  stiffness  of  a  joint  3X8 
inches  when  laid  with  flat  side  vertical  and  when  laid  with  narrow 
side  vertical. 

Prob.  66.  Find  the  thickness  of  a  white  pine  plank  of  8  feet 
span  required  not  to*  bend  more  than  ^-th  of  its  length  under  a 
head  of  water  of  20  feet. 


ART.  38.  RECAPITULATION.  67 

ART.  37.     RELATION  BETWEEN  DEFLECTION  AND  STRESS. 

Let  the  four  cases  discussed  in  Arts.  28  and  36  be  again  con- 
sidered.    For  the  strength, 


57 

=  n 

For  the  stiffness, 


W  =  n  y-     where  n  =  I,  2,  4  or 


FT  A 
W=™  -j->     where  m  =  3>  8,  48  or  76*. 

By  equating  these  values  of  W  the  relation  between  J  and 
obtained,  thus, 


=  ,  =          . 

nl*  mcE 

These  equations,  like  the  general  formula  (4)  and  (5),  are  only 
valid  when  5  is  less  than  the  elastic  limit  of  the  materials. 

This  also  shows  that  the  maximum    deflection   J  varies  as 

I2 

-   for  beams  of  the  same  material  under  the  same  unit-stress  S. 

c 

Prob.  67.  Find  the  deflection  of  a  wrought  iron  I  heavy  10 
inch  beam  of  9  feet  span  when  strained  by  a  uniform  load  up  to 
the  elastic  limit. 

Prob.  68.  Compare  the  deflections  of  a  wrought  iron  and 
wooden  beam  when  strained  to  the  elastic  limit  by  a  single  load 
at  the  middle. 

ART.  38.     RECAPITULATION. 

Let  the  length  of  the  cantilever  or  beam  be  /,  the  load  upon  it, 
whether  concentrated  or  uniform,  be  W,  the  moment  of  inertia  of 
the  constant  cross-section  about  a  horizontal  axis  through  its  cen- 
ter of  gravity  be  /,  the  shortest  distance  from  the  remotest  fiber 
to  said  axis  be  c,  the  unit-stress  at  the  elastic  limit  be  Se,  and  the 
coefficient  of  elasticity  be  E.  Then,  from  the  preceding  articles, 
the  following  table  may  be  compiled,  which  exhibits  the  most 
important  results  relating  to  both  absolute  and  relative  strength 
and  stiffness. 


68 


ON   SIMPLE    BEAMS    AND    CANTILEVERS. 


III. 


Max. 

Max. 

Max. 

Max.' 

•St 

>  £ 

Case. 

Vertical 

Bending 

Allow'ble 

Deflec- 

\D    bJO 

"w    § 

''S   c 

Shear. 

Moment. 

Load. 

tion. 

5  ^ 

Cantilever  loaded 

W 

Wl 

Sel 

,w* 

I 

I 

at  end 

Ic 

*EI 

Cantilever  loaded 

W 

y2wi 

2SeT 

*m* 

2 

2» 

uniformly 

Ic 

8  El 

3 

Simple  beam  load- 

y2W 

y^wi 

4-'— 

i  Wlz 

4 

16 

ed  at  middle 

Ic 

#EI 

Simple  beam  load- 

yw 

%wi 

85/ 

5 

8 

253 

ed  uniformly 

Ic 

384^/ 

J5 

Here  the  signs  of  the  maximum  shears  and  moments  are  omitted 
as  only  their  absolute  values  are  needed  in  computations.  Evi- 
dently the  moments  are  negative  for  the  first  and  second  cases, 
and  positive  for  the  third  and  fourth,  the  direction  of  the  curva- 
ture being  different. 

Prob.  69.  A  wooden  beam  of  breadth  b,  depth  d  and  span  x  is 
loaded  with  P  at  the  middle.  Find  the  value  of  x  so  that  rupture 
may  occur  under  the  load.  Find  also  the  value  of  x  so  that 
rupture  may  occur  by  shearing  at  the  supports. 

ART.  39.     CANTILEVERS  OF  CONSTANT  STRENGTH. 

All  cases  thus  far  discussed  have  been  of  constant  cross-section 
throughout  their  entire  length.  But  in  the  general  formula  (4) 
the  unit-stress  6"  is  proportional  to  the  bending  moment  M,  and 
hence  varies  throughout  the  beam  in  the  same  way  as  the  mo- 
ments vary.  Hence  some  parts  of  the  beam  are  but  slightly 
strained  in  comparison  with  the  dangerous  section. 

A  beam  of  uniform  strength  is  one  so  shaped  that  the  unit- 
stress  ^  is  the  same  in  all  fibers  at  the  upper  and  lower  surfaces. 
Hence  to  ascertain  the  form  of  such  a  beam  the  unit-stress  5  in 


ART.    39.  CANTILEVERS   OF    CONSTANT   STRENGTH. 


(4)  must  be  taken  as  constant  and    -  be  made  to  vary  with  M. 

The  discussion  will  be  given  only  for  the  most  important  practi- 
cal cases,  namely  those  where  the  sections  are  rectangular.     For 

these    -  equals ,  and  formula  (4)  becomes, 

c  6 

—6       -  M. 
In  this  bd2  must  vary  with  Mfor  forms  of  uniform  strength. 

For  a  cantilever  with  a  load  P  at  the  end,  M  =  Px  and  the 
equation  becomes  \Sbd2  =  Px,  in  which  P  and  5  are  constant. 
If  the  breadth  be  taken  as    constant,   d* 
varies  with  x  and  the  profile  is  that  of  a 
parabola  whose  vertex  is  at  the  load,  as 
shown  in   Fig.  26.     The   equation  of  the 

parabola  is  d2  = x    from  which  d  may 

Sb 


Elevation. 


be  found  for  given  values  of  x.  The  walk- 
ing beam  of  an  engine  is  often  made  approximately  of  this  shape. 
If  the  depth  of  the  cantilever  be  constant 
then  b  varies  directly  as  x  and  hence  the 
plan  of  the  cantilever  is  a  triangle  as  shown 
in  Fig.  27.  The  value  of  b  may  be  found 

from  the  expression  b  = 


6Px 


Plan. 


Fig.  17. 


For  a  cantilever  uniformly  loaded  with 
w  per  linear  unit  M—  y2wx*,  and  the  equation  becomes 
jrSbd*  =  y^wx*,  in  which  w  and  6"  are 
known.  If  the  breadth  be  taken  as  con- 
stant then  d  varies  as  x  and  the  elevation  is 
a  triangle,  as  in  Fig.  28,  whose  depth  at 


Elevation. 


any  point  is  d  =  x-\l~.    If  however  the 

depth  be  taken  constant,  then    b  =  ^L 

Sd* 


Fig  28. 

which  is  the   equa- 


70  ON    SIMPLE    BEAMS    AND    CANTILEVERS.  III. 

tion  of  a  parabola  whose  vertex  is  at  the 
free  end  of  the  cantilever  and  whose  axis  is 
perpendicular  to  it.  Or  the  equation  may 
be  satisfied .  by  two  parabolas  as  shown 

«-   «*»  -"  inFig'29' 

The  vertical  shear  modifies  in  practice  the  shape  of  these  forms 

near  their   ends.     For  instance,  a  cantilever  loaded  at  the  end 

p 
with  P  requires  a  cross-section  at  the  end  equal  to where  Sc  is 

the  working  shearing  strength.  This  cross-section  must  be  pre- 
served until  a  value  of  x  is  reached,  where  the  same  value  of  the 
cross-section  is  found  from  the  moment. 

The  deflection  of  a  cantilever  of  uniform  strength  is  evidently 
greater  than  that  of  one  of  constant  cross-section,  since  the  unit- 
stress  6"  is  greater  throughout.  In  any  case  it  may  be  determined 
from  the  general  formula  (5)  by  substituting  for  Mand  /their 
values  in  terms  of  x,  integrating  twice,  determining  the  constants, 
and  then  making  x  equal  to  /  for  the  maximum  value  oiy. 

For  a  cantilever  loaded  at  the  end  and  of  constant  breadth,  as 
in  Fig.  26,  formula  (5)  becomes, 

dy  _  i  zPx 2 


Integrating   this   twice   and   determining   the    constants,   as    in 
Art.  34,  the  equation  of  the  elastic  curve  is  found  to  be, 

/-IT-T.T 

In  this  make  x  =  /,  and  substitute  for  5  its  value    __,  where 

Uf 
dr  is  the  depth  of  the  wall.     Then, 

8P/3 


which  is  double  that  of  a  cantilever  of  constant  cross-section, 


ART.    40.  SIMPLE    BEAMS    OF    UNIFORM    STRENGTH.  /I 

For  a  cantilever  loaded  at  the  end  and  of  constant  depth, 
formula  (5)  becomes, 

2$ 
~Ed 


By  intergrating  this  twice  and  determining  the  constants  as 
before,  the  equation  of  the  elastic  curve  is  found,  from  which  the 
deflection  is, 

6P/3 

Ebd* 
which  is  fifty  per  cent  greater  than  for  one  of  uniform  section. 

Prob.  70.  A  cast-iron  cantilever  of  uniform  strength  is  to  be 
4  feet  long,  3  inches  in  breadth  and  to  carry  a  load  of  1  5  ooo 
pounds  at  the  end.  Find  the  proper  depths  for  every  foot  in 
length,  using  3  ooo  pounds  per  square  inch  for  the  horizontal 
unit-stress,  and  4  ooo  pounds  per  square  inch  for  the  shearing 
unit-stress. 

ART.  40.     SIMPLE   BEAMS  OF  UNIFORM  STRENGTH. 

In  the  same  manner  it  is  easy  to  deduce  the  forms  of  uniform 
strength  for  simple  beams  of  rectangular  cross-section. 

For  a  load  at  the  middle  and  breadth  constant,  M=  y?,Px,  and 

i>P 
hence,  ^Sbd2  =  y^Px.      Hence  d*  =  ^-*,    from  which  values 

Ov 

of  d  may  be  found  for  assumed  values  of  x.  Here  the  profile  of 
the  beam  will  be  parabolic,  the  vertex  being  at  the  support,  and 
the  maximum  depth  under  the  load. 

For  a  load  at  the  middle  and  depth  constant,  M  =  y^Px  as 

*p 
before.    Hence  b  =  ~^~r2^,  and  the  plan  must  be   triangular   or 

lozenge  shaped,  the  width  uniformly  increasing  from  the  support 
to  the  load. 

For  a  uniform  load  and  constant  breadth,  M=  y2wlx  —  ^w;r2, 
and  hence,  d*=  ~^r(lx  —  *2),  and  the  profile  of  the  beam  must 
be  elliptical,  or  preferably  a  half-ellipse, 


72  ON    SIMPLE    BEAMS    AND    CANTILEVERS.  III. 

T.W 

For  a  uniform  load  and  constant  depth,  b  =  ~^T^X  —  x*}  an<3 

hence  the  plan  should  be  formed  of  two  parabolas  having  their 
vertices  at  the  middle  of  the  span. 

The  figures  for  these  four  cases  are  purposely  omitted,  in  order 
that  the  student  may  draw  them  on  the  margin.  In  the  same 
manner  as  in  the  last  Article,  it  can  be  shown  that  the  deflection 
of  a  beam  of  uniform  strength  loaded  at  the  middle  is  double  that 
of  one  of  constant  cross-section  if  the  breadth  is  constant,  and  is 
one  and  one-half  times  as  much  if  the  depth  is  constant. 

Prob.  71.  Find  the  deflection  of  a  steel  spring  of  constant  depth 
and  uniform  strength  which  is  6  inches  wide  at  the  middle,  52 
inches  long,  and  loaded  at  the  middle  with  600  pounds,  the  depth 
being  such  that  this  load  strains  the  material  to  one-half  its  elastic 
limit. 


ART.   41.  GENERAL    PRINCIPLES.  73 


CHAPTER  IV. 

ON  RESTRAINED  BEAMS  AND  ON  CONTINUOUS  BEAMS. 

ART.  41.  GENERAL  PRINCIPLES. 

A  restrained  beam  is  one  whose  ends  are  fastened  in  walls,  or 
so  arranged  that  the  tangents  to  the  elastic  curve  at  the  ends  al- 
ways remain  horizontal.  The  simplest  case  of  a  restrained  beam 
is  a  cantilever  with  one  end  fixed  and  the  other  free.  Two  other 
common  cases  are  a  beam  fixed  at  one  end  and  supported  at  the 
other,  and  a  beam  fixed  at  both  ends. 

A  continuous  beam  is  one  supported  upon  several  points  in  the 
same  horizontal  plane.  A  simple  beam  may  be  regarded  as  a 
particular  case  of  a  continuous  beam  where  the  number  of  sup- 
ports is  two.  The  ends  .of  a  continuous  beam  are  said  to  be  free 
when  they  overhang,  supported  when  they  merely  rest  on  abut- 
ments, and  restrained  when  they  are  horizontally  fixed  in  walls. 

The  general  principles  of  the  preceding  Chapter  hold  good  for 
all  kinds  of  beams.  If  a  plane  be  imagined  to  cut  any  beam  at 
any  point  the  laws  of  Arts.  15  and  18  apply  to  the  stresses  in  that 
section.  The  resisting  shear  and  the  resisting  moment  for  that 
section  have  the  values  deduced  in  Art.  19,  and  as  in  Art.  20  the 
two  fundamental  formulas  for  investigation  are, 
(3)  S<A  =  V, 

4  ££    =  M. 


74  RESTRAINED    BEAMS   AND    CONTINUOUS    BEAMS.  IV. 

Here  ^  is  the  vertical  shearing  unit-stress  in  the  section,  and 
5"  is  the  horizontal  tensile  or  compressive  unit-stress  on  the  fiber 
most  remote  from  the  neutral  axis ;  c  is  the  shortest  distance  from 
that  fiber  to  that  axis  ;  /  the  moment  of  inertia,  and  A  the  area  of 
the  cross-section.  V"\s  the  vertical  shear  of  the  external  forces 
on  the  left  of  the  section,  and  M  is  the  bending  moment  of  those 
forces  with  reference  to  a  point  in  the  section.  For  any  given 
beam  evidently  Ss  and  6"  may  be  found  for  any  section  as  soon 
as  Fand  Mare  known. 

The  general  equation  of  the  elastic  line,  deduced  in  Art.  33,  is 
also  valid  for  all  kinds  of  beams.  It  is, 

(5)  »?L-*L 

dx*~  El 

where  x  is  the  abscissa  and  y  the  ordinate  of  any  point  of  the 
elastic  curve,  M  being  the  bending  moment  for  that  section,  and 
E  the  coefficient  of  elasticity  of  the  material. 

The  vertical  shear  V  is  the  algebraic  sum  of  the  external  forces 
on  the  left  of  the  section,  or,  as  in  Art.  16, 

V  =  Reactions  on  left  of  section  minus  loads  on  left  of  section. 
For  simple  beams  and  cantilevers  the  determination  of  FTor  any 
special  case  was  easy,  as  the  left  reaction  could  be  readily  found  for 
any  given  loads.  For  restrained  and  continuous  beams,  however, 
it  is  not,  in  general,  easy  to  find  the  reactions,  and  hence  a  differ- 
ent method  of  determining  V  is  necessary.  Let  Fig.  30  repre- 
sent one  span  of  a  continuous  or  restrained  beam.  Let  V  be  the 

p jp ^  vertical  shear   for   any 

j, -JB-— f       | section  at  the  distance 

)  j  \     x  from  the  left  support, 

k"  and  V  the  vertical  shear 

Flg30.  .  .          . 

at   a   section    infinitely 

near  to  the  left  support.  Also  let  IP  denote  the  sum  of  all  the 
concentrated  loads  on  the  distance  x,  and  wx  the  uniform  load. 


ART.    41.  GENERAL    PRINCIPLES.  75 

Then  because  V  is  the  algebraic  sum  of  all  the  vertical  forces  on 
its  left,  the  definition  of  vertical  shear  gives, 

(6)  V=V—IP—wx. 

H-ence  Fcan  be  determined  as  soon  as  V  is  known. 

The  bending  moment  M  is  the  algebraic  sum  of  the  moments 
of  the  external  forces  on  the  left  of  the  section  with  reference  to 
a  point  in  that  section,  or,  as  in  Art.  17, 

M  =  Moments  of  reactions  minus  moments  of  loads. 

For  the  reason  just  mentioned  it  is  in  general  necessary  to  de- 
termine M  for  continuous  and  restrained  beams  by  a  different 
method.  Let  M'  denote  the  bending  moment  at  the  left  support 
of  any  span  as  in  Fig.  30,  and  M"  that  at  the  right  support,  while 
M  is  the  bending  moment  for  any  section  distant  x  from  the  left 
support.  Let  P  be  any  concentrated  load  upon  the  space  x  at  a 
distance  kl  from  the  left  support,  k  being  a  fraction  less  than 
unity,  and  let  w  be  the  uniform  load  per  linear  unit.  Then,  be- 
cause M  is  the  algebraic  sum  of  all  the  moments  of  the  external 
forces  on  its  left,  the  definition  of  bending  moment  gives, 
(7)  M  =  M  +  Vx  —  IP  (x  —  kl}  —  y2wx*. 

Hence  Mmay  be  found  for  any  section  as  soon  as  V  and  M1 
are  known. 

The  relation   between  the  bending  moment  and  the  vertical 
shear  at  any  section  is  interesting  and  important.     At  the  section 
x  the  moment  is  M  and  the  shear  is  V.     At  the  next  consecutive 
section  x  -f  dx  the  moment  is  M-\-  dM,  which  may  also  be  ex- 
pressed by  M-\-  Vdx.     Hence, 
v=dM 
dx 

This  may  be  proved  otherwise  by  differentiating  (7)  and  compar- 
ing with  (6). 

The  vertical  shear  V  at  the  support  may  be  easily  found  if  the 
bending  moments  M'  and  M"  be  known.  Thus  in  equation 


76 


RESTRAINED    BEAMS   AND    CONTINUOUS    BEAMS. 


IV. 


(7)  make  ;tr=  /,  then  M  becomes  M" ',  and  hence, 
(8)  V^tZ+VL 


The  whole  problem  of  the  discussion  of  restrained  and  contin- 
uous beams  hence  consists  in  the  determination  of  the  bending 
moments  at  the  supports.  When  these  are  known  the  values  of 
Mand  V  may  be  determined  for  every  section,  and  the  general 
formulas  (3),  (4)  and  (5)  be  applied  as  in  Chapter  III,  to  the  in- 
vestigation of  questions  of  strength  and  deflection.  The  formulas 
(6),  (7)  and  (8)  apply  to  simple  beams  and  cantilevers  also.  For 
simple  beams  M1  =  M"  =  o,  and  V  =  R  since  there  is  no  re- 
straint at  the  ends.  For  cantilevers  M1  =  o  for  the  free  end,  and 
M"  is  the  moment  at  the  wall. 

Prob.  72.  A  bar  of  length  2/  and  weighing  w  per  linear  unit  is 
supported  at  the  middle.  Apply  formulas  (6)  and  (7)  to  the  state- 
ment of  general  expressions  for  the  moment  and  shear  at  any 
section  on  the  left  of  the  support,  and  also  at  any  section  on  the 
right  of  the  support. 

ART.  42.      BEAMS   OVERHANGING   ONE  SUPPORT. 

A  cantilever  has  its  upper  fibers  in  tension  and  the  lower  in 

compression,   while    a 


simple  beam  has  its 
upper  fibers  in  com- 
pression and  the  lower 
in  tension.  Evidently 
a  beam  overhanging 
one  support,  as  in  Fig. 
3 1 ,  has  its  overhanging 
part  in  the  condition 
of  a  cantilever,  and  the 
part  near  the  other  end 
in  the  condition  of  a 
simple  beam.  Hence  there  must  be  a  point  i  where  the  stresses 


Moments. 


Shears. 


Fig.  31. 


Pi 


ART.    42.  BEAMS    OVERHANGING    ONE   SUPPORT.  77 

change  from  tension  to  compression,  and  where  the  curvature 
changes  from  positive  to  negative.  This  point  i  is  called  the  in- 
flection point ;  it  is  the  point  where  the  bending  moment  is  zero. 
Since  the  beam  has  but  two  supports  its  reactions  may  be  found, 
as  in  Art.  14,  and  the  entire  investigation  be  made  by  the  prin- 
ciples of  the  last  Chapter. 

Consider  a  beam  loaded  uniformly  with  w  per  linear  unit.     Let 
/  be  the  distance  between  the  supports  and  m  the  length  of  the 
overhanging  portion.     Let  the  left  reaction  be  RT  and  the  right 
R2.     Then,  for  any  section  distant  x  from  the  left  support, 
When  x  is  less  than  /,  When  x  is  greater  than  /, 

V  =  Rt  —  wx,  V=R^  +  R2  —  wx, 

M  =  RS  —  y2wx*.  M=Rs+ R2  (x — /)—  wx\ 

The  curves  corresponding  to  these  equations  are  shown  on 

Fig.  31.     The  shear  curve  consists  of  two  straight  lines  ;    V  =.  Rz 

£ 
when  x  =  o,    V  =  o   when  x  =  — '  ;   at  the   right  support 

w 

V=  RT  —  wl  from  the  first  equation,  and  V—  R^  +  R2  —  w/from 
the  second ;  V  =  o  when  x  =  /  -f-  m.  The  moment  curve  consists 
of  two  parts  of  parabolas  ;  M  —  o  when  x  =  o,  M  is  a  maximum 

]T>  TT> 

when   x  =  -—,  M=  o  at  the  inflection  point  where  x—  — -,   M 

w  w 

has  its  negative  maximum  when  x  =  /,  and  M  =  o  when 
x  =  I  +  m.  The  diagrams  show  clearly  the  distribution  of 
shears  and  moments  throughout  the  beam. 

For  example,  if  /  =  20  and  m  =  10  the  reactions  are  found  to 
be  Rz  =j.$w  and  R2  —  22. .$w.  Then  the  point  of  zero  shear  or 
maximum  moment  is  at  x  =  7.5,  the  inflection  point  at  x  =  15, 
the  maximum  shears  are  +  7.5^,  —  \2.$w  and  +  low,  and 
the  maximum  moments  are  -f-  56.25^  and  —  5ow.  The  relative 
values  of  the  two  maximum  moments  depend  on  the  ratio  of  m  to  /. 
If  this  ratio  be  zero  there  is  no  overhanging  part  and  no  negative 
moment,  and  the  beam  is  a  simple  one.  If  m  —  y^l  the  case  is 


78        RESTRAINED  BEAMS  AND  CONTINUOUS  BEAMS.       IV. 

that  just  discussed.     If  m  —  /  the  beam  becomes  a  cantilever  sup- 
ported at  the  middle. 

After  having  thus  found  the  maximum  values  of  V  and  M  the 
beam  may  be  investigated  by  the  application  of  formulas  (3)  and 
(4)  in  the  same  manner  as  a  cantilever  or  simple  beam.  By  the 
use  of  formula  (5)  the  equation  of  the  elastic  curve  between  the 
two  supports  is  found  to  be, 

2^Efy  =  4Rt  (x*  —  t*x)—w(x*  —  fix). 

From  this  the  maximum  deflection  for  any  particular  case  may  be 

dv 
determined  by  putting  —  equal  to  zero,  solving  for  x,  and  then 

dx 
finding  the  corresponding  value  of y. 

Prob.  73.  Find  the  ratio  of  m  to  /so  that  the  maximum  posi- 
tive moment  may  equal  numerically  the  maximum  negative 
moment. 

Prob.  74.  A  light  1 2-inch  /beam  25  feet  long  is  used  as  a  floor 
beam  in  a  bridge  with  one  sidewalk,  the  distance  between  the  sup- 
ports being  20  feet.  Find  its  factor  of  safety  when  the  whole 
beam  is  loaded  with  I  2OO.pounds  per  linear  foot,  and  also  when 
only  the  20  feet  roadway  is  loaded. 

ART.   43.     BEAMS    FIXED   AT   ONE    END   AND   SUPPORTED    AT 

THE   OTHER. 

When  a  beam  is  fixed  horizontally  in  a  wall  at  one  end  while 
the  other  end  is  merely  supported,  the  restraint  at  the  fixed  end 
causes  the  distribution  of  moments  to  be  similar  to  that  of  a  beam 
with  one  overhanging  end.     The  reaction  at  the  supported  end 
cannot  be  found  by  the  principles  of 
j  |  i  |  i        statics  as  in  Art.   14,  but  must  be 
,  P"7"      determined  by  the  help  of  the  equa- 
tion of  the  elastic  curve. 


y- 

LrrnTTTTTTnv 


\|l  Case  I. — For  a  uniform  load  over 

the  whole  beam  let  R  be  the   left 
reaction  as  in  Fig.  32.     Then  for  any  section  the  bending  moment 


ART.  43.  BEAMS  FIXED  AND  SUPPORTED.  79 

is  Rx  —  y2wxz.     Hence  the  differential  equation  of  the  elastic 
curve  is, 

EI*L  =  Rx—  V^wx\ 

dx* 

Integrate  this  once  and  determine  the  constant  from  the  necessary 

dv 

condition  that  —  —  =  o  when  x  =  I.      Integrate   again  and  find 
dx 

the  constant  from  the  fact  thatjy  =  o  when  x  =  o.     Then, 


Here  also^  =  o  when  x  —  /,  and  therefore  R  = 

The  moment  at  any  point  now  is  M  =  y#wlx  —  y2wxz,  and 
by  placing  this  equal  to  zero  it  is  seen  that  the  point  of  inflection 
is  at  x  =  y^l.  By  the  method  of  Art.  23  it  is  found  that  the  max- 
imum moments  are  +  _L  wl*  and  —  iwP,  and  that  the  distribu- 
tion of  moments  is  as  represented  in  Fig.  3.2 

The  point  of  maximum  deflection  is  found  by  placing  --L.   equal 

dx 
to  zero  and  solving  for  x.     This  gives  8x3  —  glx*  -\-  fi  =  o,  one 

root  of  which  is  x  =  +  0.42157,  and  this  inserted  in  the  value 
of  y  gives, 

wfo 
A  =  0.0054  _  , 

for  the  value  of  the  maximum  deflection. 

Case  II.  —  For  a  load  at  the  middle  it  is  first  necessary  to  con- 
sider that  there  are  two  elastic  curves  having  a  common  ordinate 
and  a  common  tangent  under  the  load,  since  the  expressions  for 
the  moment  are  different  on  opposite  sides  of  the  load.  Thus 
taking  the  origin  as  usual  at  the  supported  end, 

On  the  left  of  the  load, 


(6)  El-  =  */2Rx  +  C, 

dx 


8O        RESTRAINED  BEAMS  AND  CONTINUOUS  BEAMS.       IV. 

On  the  right  of  the  load  the  similar  equations  are, 


(by      Ei-j-x 

(cy  Ely  =  \Rx*  -  IP**  +  KPb*  +  Cs  +  Q 

To  determine  the  constants  consider  in  (c)  that  y  =  o  when  x  =  o 
and  hence  that  C3  =  o.  In  (b)'  the  tangent  -^  =  o  when  x  =  I 

and  hence  C2  =  —  y*.Rl.  Since  the  curves  have  a  common  tangen 
under  the  load  (b)  =  (b}'  for  x  —  ^/,  and  thus  the  value  of  Cz  is 
found.  Since  the  curves  have  a  common  ordinate  under  the  load 
(c)  =  (c}'  when  x  =  ^/,  and  thus  C4  is  found.  Then, 


Rx*       Px*        Plx*        Rl*x        PI* 

-sr  ---6-+—    -F+ir 

From  the  second  of  these  the  value  of  the  reaction  is  R  =  ~P. 

The  moment  on  the  left  of  the  load  is  now  M  =  -^-  Px,  and 
that  on  the  right  M  =  —  £  Px  +  l/2Pl.  The  maximum  positive 
moment  obtains  at  the  load  and  its  value  is  —a  PI.  The  maximum 
negative  moment  occurs  at  the  wall,  and  its  value  is  ~  PL  The 
inflection  point  is  at  x  =  -^  I.  The  deflection  under  the  load  is 
readily  found  from  (c  )  by  making  x  =  *-/.  The  maximum  de- 
flection occurs  at  a  less  value  of  x,  which  may  be  found  by 

dy 

putting  ——  =  o. 
dx 

Case  III.  For  a  load  at  any  point  whose  distance  from  the  left 
support  is  kl,  the  following  results  may  be  deduced  by  a  method 
exactly  similar  to  that  of  the  last  case. 

Reaction  at  supported  end  =  /^/^2  —  3^  -f-  k$). 

Reaction  at  fixed  end  =  Y^P^k  —  fc\ 

Maximum  positive  moment  =  y^Plk(2  —  3/£3  +  k$). 

Maximum  negative  moment  =  y^Pl(k  —  £3). 


ART.   45.  BEAMS   FIXED   AT   BOTH    ENDS.  8 1 

The  absolute  maximum  deflection  occurs   under  the  load  when 
k  =  0.42  2/. 

Prob.  75.  Find  the  position  of  load  P  which  gives  the  maxi- 
mum positive  moment.  Find  also  the  position  which  gives  the 
maximum  negative  moment. 

ART.  44.     BEAMS  OVERHANGING  BOTH  SUPPORTS. 

When  a  beam  overhangs  both  supports,  as  in  Fig.  33,  there 
will  be  a  negative  moment  at 
each  support  and,  in  general, 
two  inflection  points  and  a 
positive  moment.  The  rela- 
tive values  of  these  will  depend 
upon  the  lengths  of  the  over- 
hanging parts.  If  these  lengths  m  and  n  be  equal,  the  reactions 
and  negative  moments  will  be  equal  and  the  two  halves  of  the 
beam  be  symmetrically  strained  by  a  uniform  load.  In  any  case, 
whatever  be  the  nature  of  the  load,  the  reactions  may  be  found 
by  Art.  14  and  the  maximum  vertical  shears  and  bending  mo- 
ments be  determined  by  the  methods  of  the  last  Chapter. 

Prob.  76.  If  m  =  n  in  Fig.  33,  find  the  ratio  of  m  to  I  in  order 
that  the  maximum  positive  moment  may  numerically  equal  the 
maximum  negative  moments. 

Prob.  77.  A  bridge  with  two  sidewalks  has  a  wooden  floor 
beam  14  X  15  inches  and  30  feet  long,  the  distance  between  sup- 
ports being  20  feet  and  each  sidewalk  5  feet.  Find  its  factor  of 
safety  under  a  uniformly  distributed  load  of  29  ooo  pounds. 

ART.  45.     BEAMS  FIXED  AT  BOTH  ENDS. 

Case  I.  For  a  uniform  load  it  is  evident  that  the  bending 
moments  at  the  supports  are  equal.  Then,  for  any  section  x, 
formula  (7)  becomes, 

M=Mf  +  Rx—  y2wx*, 
in  which  M'  is  the  unknown  moment  at  the  left  support.     By 


82        RESTRAINED  BEAMS  AND  CONTINUOUS  BEAMS.       IV. 

inserting  this  in  (5)  and  integrating  twice,  making  —  =  o  when 

dx 

i  f  i  |  u-a:-»!  I     I,       x  =  o  and  also  when  .r  =  /, 

h-1-     the  value  of  M1  is  found  to 
'  w/2 

-  -£  and  the  equation 

K  Flg.34  N|  of  the  elastic  curve  is, 

2^EIy  =  w(  —  l*x*  +  2/a*  _  A4). 
From  this  the  maximum  deflection  is  found  to  be, 


The  inflection  points  are  located  by  making  M=  o.     This  gives 


x  =  ^2/±/^|      .    The  maximum   positive    moment   is   at   the 

middle  and  its  value  is  —  . 
24 

Case  II.  For  a  load  at  the  middle,  the  bending  moment  between 
p  the  left  end  and  the  load  is, 

M  =  M  +  y2Px, 
and  in  a  similar  manner  to  that  of 
the  last  case  it  is  easy  to  find  that 
the  maximum  negative  moments 


Fig35.  are  1/&PI,  that  the  maximum  posi- 

*3 
tive  moments  are  1APL  and  that  the  maximum  deflection  is 


Case  III.  For  a  load  P  at  a  distance  kl  from  the  left  end,  a 
method  similar  to  that  of  Case  II,  Art.  43,  may  be  followed.  Let 
M  and  R'  denote  the  unknown  moment  and  reaction  at  the  left 
end  and  M"  and  R"  those  for  the  right  end.  Then, 

On  the  left  of  the  load, 

(a)  £70  =  M  +  R'x, 

(b)  £1^=  M>x+  y2R>x\ 

(c)  Ely  =  y2Mx*  +  \R'x\ 


ART.  46.    COMPARISON  OF  RESTRAINED  AND  SIMPLE  BEAMS.  83 

On  the  right  of  the  load, 

(a)r      EId-~  =  M*  +  R'x  +  P(x  —  kl) 


(c)'  Ely  = 

The  constants  of  integration  are  here  readily  determined  since 

both  -r  .    and  y  become  zero  for  x  =  o  and  for  x  =  I.     Then 
dx 

if  x  —  kl,  the  tangent  (6)  is  equal  to  (£)',  and  the  ordinate  (c]  to 
(cy.     This  gives  two  equations  from  which  the  values  of  M1  and 
R'  are  found.     Thus  the  following  results  are  deduced, 
Reaction  at  left  end        =  P(i  —  3/£2  +  2^), 
Reaction  at  right  end    =  Pk*  (3  —  2k\ 
Moment  at  left  end        =  —  Plk  (i  —  2/£  +  fr), 
Moment  at  right  end     =  —  Plk2  (i  —  k\ 
Moment  under  load        =  +  Plk2  (2  —  $k  +  2/£2). 
If  k  =  X  ^e  load  is  at  the  middle  and  these  results  reduce  to 
the  values  found  in  Case  II. 

Prob.  78.  What  wrought  iron  I  beam  is  required  for  a  span  of 
24  feet  to  support  a  uniform  load  of  40  ooo  pounds,  the  ends 
being  merely  supported  ?  What  one  is  needed  when  the  ends 
are  fixed  ? 

ART.  46!  COMPARISON  OF  RESTRAINED  AND  SIMPLE  BEAMS. 
As  the  maximum  moments  for  restrained  beams  are  less  than 
for  simple  beams  their  strength  is  relatively  greater.  This  was 
to  be  expected  since  the  restraint  produces  a  negative  bending 
moment  and  lessens  the  deflection  which  would  otherwise  occur. 
The  comparative  strength  and  stiffness  of  cantilevers  and  simple 
beams  is  given  in  Art.  38.  To  these  may  now  be  added  four 
cases  from  Arts.  43  and  45,  and  the  following  table  be  formed, 
in  which  W  represents  the  total  load,  whether  uniform  or 
concentrated. 


84 


RESTRAINED  BEAMS  AND  CONTINUOUS  BEAMS. 


IV. 


Beams  of  Uniform  Cross-section. 

Maximum 
Moment. 

Maximum 
Deflection. 

Relative 
Strength. 

Relative 

Stiffness. 

Cantilever,  load  at  end 

Wl 

LWP2j 

I 

! 

*  El 

Cantilever;  uniform  load 

-Wl 

JVfr 

2 

2- 

8  El 

3 

Simple  beam,  load  at  middle 

-Wl 

•m* 

4 

16 

4 

48  p  T 

Simple     beam      uniformly 

loaded 

-  Wl 

5 

8 

25-3 

8 

384^7 

5 

Beam  fixed  at  one  end,  sup- 

ported at  other,  load  near 

middle 
Beam  fixed  at  one  end,  sup- 

0.174^ 

°.°99§ 

6 

34 

ported  at  other,  uniform 

load 

\Wl 

0.0054^ 

8 

62 

Beam    fixed   at  both  ends, 

load  at  middle 

\Wl 

=5^7 

8 

64 

Beam    fixed  at  both  ends, 

Jl/h 

uniform  load 

12    ^^ 

!      Wl* 

1  2 

128 

**EI 

This  table  shows  that  a  beam  fixed  at  both  ends  and  uniformly 
loaded  is  one  and  one-half  times  as  strong  and  five  times  as  stiff 
as  a  simple  beam  under  the  same  load.  The  advantage  of  fixing 
the  ends  is  hence  very  great. 

Prob.  79.  Find  the  deflection  of  a  I  9-inch  beam  of  6  feet  span 
and  fixed  ends  when  strained  so  that  the  tensile  and  compressive 
stresses  at  the  dangerous  section  are  14  ooo  pounds  per  square 
inch. 

ART.  47.      PROPERTIES  OF  CONTINUOUS  BEAMS. 
The  theory  of  continuous  beams  presented  in  the  following 
pages  includes  only  those  with  constant  cross-section  having  the 
supports  on  the  same  level,  as  only  such  are  used  in  engineering 


ART.  47.      PROPERTIES  OF  CONTINUOUS  BEAMS.  85 

constructions.  Unless  otherwise  stated,  the  ends  will  be  supposed 
to  simply  rest  upon  their  supports,  so  that  there  can  be  no  mo- 
ments at  those  points.  Evidently  then  the  end  spans  are  some- 
what in  the  condition  of  a  beam  with  one  overhanging  end  and 
the  other  spans  somewhat  in  the  condition  of  a  beam  with  two 

overhanging 

fe  A5       I, i, i,  14 

ends.         At      {  \ 

each     inter-          f*  ?*  ~~f»  f^  f* 

mediate  sup-  LnTTTTrrv          ..xrrrrrx         .xmv   I 

port  there  is  ^W  ^f 

a      negative  X       Fi&3& 

moment,  and 

the  distribution  of  moments  throughout  the  beam  will  be  as  rep- 
resented in  Fig.  36. 

As  shown  in  Art.  41  the  investigation  of  a  continuous  beam 
depends  upon  the  determination  of  the  bending  moments  at  the 
supports.  In  the  case  of  Fig.  36  these  moments  being  those  at 
the  supports  2,  3  and  4,  may  be  designated  M2,  M3  and  M^.  Let 
V^  Vz,  V^  and  V^  denote  the  vertical  shear  at  the  right  of  each 
support.  The  first  step  is  to  find  the  moments  M2,  M3  and  M^. 
Then  from  formula  (8)  the  values  of  Vt,  V2,  V^  and  V^  are  found, 
and  thus  by  formula  (7)  an  expression  for  the  bending  moment 
in  each  span  may  be  written,  from  which  the  maximum  positive 
moments  may  be  determined.  Lastly,  by  formula  (4)  the  strength 
of  the  beam  may  be  investigated  and  by  (5)  its  deflection  at  any 
point  be  deduced. 

For  example,  let  the  beam  in  Fig.  36  be  regarded  as  of  four 
equal  spans  and  uniformly  loaded  with  w  pounds  per  linear  unit. 
By  a  method  to  be  explained  in  the  following  articles  it  may  be 
shown  that  the  bending  moments  at  the  supports  are, 

M2=  —  ^wl2,         M3=  —  ^wl2,         M^  =  —  ^wl2,. 
From  formula  (8)  the  vertical  shears  at  the  right  of  the  supports 
are, 


86  RESTRAINED    BEAMS   AND    CONTINUOUS    BEAMS.  IV. 

F;  ==  g  wl,       V2  =  ^  wl,       V=  I  wl,      etc. 
And  from  (6)  those  on  the  left  of  the  supports  are  found  to  be, 
F/  =  ->/,         F2'=-|W/,         ra'  =  ->l,     etc. 
From  formula  (7)  the  general  moments  now  are, 


For  first  span,         M  =      ^  wlx  — 
For  second  span,   M  =  —  ~  wl2  + 


-  ^vx'2. 


>&->*•, 


For  third  span,       M  =  —  ^  wl2  - 

For  fourth  span,  M  =  —  ~  wfi  -f-  ^  wlx  —  ^  wx2. 
From  each  of  these  equations  the  inflection  points  may  be  found 
by  putting  M=o,  and  the  point  of  maximum  positive  moment 

by  putting =  o.     The  maximum  positive  moments  are, 


For  any  particular  case  the  beam  may  now  be  investigated  by 
formulas  (3)  and  (4). 

The  reactions  at  the  supports  are  not  usually  needed  in  the 
discussion  of  continuous  beams,  but  if  required  they  may  easily 
be  found  from  the  adjacent  shears.  Thus  for  the  above  case, 

Rt  =     o    +  Vt  =  ~  wl, 

R2=  V;+  V9=*wl, 

R3=  V2*+  V3  =  £u>t,     etc, 
and  the  sum  of  these  is  equal  to  the  total  load  ^wl. 

The  equation  of  the  elastic  curve  in  any  span  is  deduced  by 
inserting  in  (5)  for  M  its  value  and  integrating  twice.  When 

x  =  o,  the  tangent  -- -    is  the  tangent  of  the  inclination  at  the 
dx 

left  support  and  when  x  =  I  it  is  the  tangent  of  the  inclination  at 
the  right  support.  When  x  =  o  and  also  when  x  =  I  the  ordi- 
nate  y  =  o,  and  from  these  conditions  the  two  unknown  tangents 
may  be  found.  In  general  the  maximum  deflection  in  any  span 


ART.  48.       THE  THEOREM  OF  THREE  MOMENTS.  87 

of  a  continuous  beam  will  be  found  intermediate  in  value  between 
those  of  a  simple  beam  and  a  restrained  beam. 

In  the  following  pages  continuous  beams  will  only  be  investi- 
gated for  the  case  of  uniform  load.  The  lengths  of  the  spans 
however  may  be  equal  or  unequal,  and  the  load  per  linear  foot 
may  vary  in  the  different  spans. 

Prob.  80.  In  a  continuous  beam  of  three  equal  spans  the  nega- 
tive bending  moments  at  the  supports  are  ^wlz.  Find  the  in- 
flection points,  the  maximum  positive  moments  and  the  reactions 
of  the  supports. 

ART.  48.     THE  THEOREM  OF  THREE  MOMENTS. 
Let  the  figure  represent  any  two  adjacent  spans  of  a  continu- 
ous beam  whose  lengths  are  /'  and  I"  and  whose  uniform  loads 
per  linear  foot  are  w'  and  w"  respectively.  Let  M7,  M"  and  M"' 
represent  the  three  un- 
known moments  at  the          K%%%%%3%%^^ 

^^^^^^^^%%^^^^^^^^$$$$$^$$^ 

supports.  Let  V  and      }  \ 

V"   be    the     vertical  |y*  I V* . 

shears  at  the  right  of  Tig. 37 

the  first  and  second  supports.     Then,  for  any  section  distant  x 

from  the  left  support  in  the  first  span,  the  moment  is, 

M=  M1  +  Vx—  y2wx\ 

If  this  be  inserted  in  the  general  formula  (5)  and  integrated  twice 
and  the  constants  determined  by  the  condition  that  y  =  o  when 
x  =  o  and  also  when  x  =  I,  the  value  of  the  tangent  of  the  angle 
which  the  tangent  to  the  elastic  curve  at  any  section  in  the  first 
span  makes  with  the  horizontal  is  found  to  be, 

dy        \2M\2x  —  I'}  -f  4 V^x*  —  /'2)  —  w'fa*  —  I'*} 

dx  2AJELI 

Similarly  if  the  origin  be  taken  at  the  next  support  the  value  of 
the  tangent  of  inclination  at  any  point  in  the  second  span  is, 

dy         \2M"(2x  —  /")  +  4V"($x2  —  ///2)  —  w"(4x*  —  I"*) 


88  RESTRAINED    BEAMS   AND   CONTINUOUS    BEAMS.  IV. 

Evidently  the  two  curves  must  have  a  common  tangent  at  the 
support.  Hence  make  x  =  I'  in  the  first  of  these  and  x  =  o  in 
the  second  and  equate  the  results,  giving, 

i-zM'l'  +  8  FV2  —  3w'/'3  =  —  i2M"l"  —  4F"/"2  +  w"l"*. 
Let  the  values  of  V  and  V"  be  expressed  by  (8)  in  terms  of  M1  ', 
M"  and  M"f,  and  the  equation  reduces  to, 

(9)        M'F  +  2M»(P 


4  4 

which  is  the  theorem  of  three  moments  for  continuous   beams 
uniformly  loaded. 

If  the  spans  are  all  equal  and   the   load  uniform   throughout, 
this  reduces  to  the  simpler  form, 


In  any  continuous  beam  of  s  spans  there  are  s  -j-  i  supports 
and  s  —  I  unknown  bending  moments  at  the  supports.  For 
each  of  these  supports  an  equation  of  the  form  of  (9)  may  be 
written  containing  three  unknown  moments.  Thus  there  will  be 
stated  s  —  I  equations  whose  solution  will  furnish  the  values  of 
the  s  —  i  unknown  quantities. 

Prob.  8  1.  A  simple  beam  of  oak  one  inch  square  and  15  inches 
long  is  uniformly  loaded  with  100  pounds.  Find  the  angle  of 
inclination  of  the  elastic  curve  at  the  supports. 

ART.  49.     CONTINUOUS  BEAMS  WITH  EQUAL  SPANS. 

Consider  a  continuous  beam  of  five   equal  spans    uniformly 

loaded.     Let  the  supports  beginning  on  the  left  be  numbered 

I,  2,  3,  4,  5  and  6.       From  the  theorem    of  three   moments    an 

equation  may  be  written  for  each  of  the  supports  2,  3,  4,  and  5  ; 

thus, 

M  +    M  +  M  = 


ART.  49.       CONTINUOUS  BEAMS  WITH  EQUAL  SPANS.  89 

Since  the  ends  of  the  beam  rest  on  abutments  without  restraint 
Ml  =  M6  =  o.  Hence  the  four  equations  furnish  the  means  of 
finding  the  four  moments  M2,  M^  M^  Me.  The  solution  may  be 
abridged  by  the  fact  that  M2  =  MS,  and  M^  =  M^,  which  is  evi- 
dent from  the  symmetry  of  the  beam.  Hence, 


2          5       —          ,  3          4 

From  formula  (8)  the  shears  at  the  right  of  the  supports  are, 
K=iX  V,  =  l^  V3=^wl,      etc. 

From  (6)  the  bending  moment  at  any  point  in  any  span  may  now 
be  found  as  in  Art.  47,  and  by  (3),  (4)  and  (5)  the  complete  inves- 
tigation of  any  special  case  may  be  effected. 

In  this  way  the  bending  moments  at  the  supports  for  any  num- 
ber of  equal  spans  can  be  deduced.  The  following  triangular 
table  shows  their  values  for  spans  as  high  as  seven  in  number. 
In  each  horizontal  line  the  supports  are  represented  by  squares 
in  which  are  placed  the  coefficients  of  wl2.  For  example,  in  a 
beam  of  3  spans  there  are  four  supports  and  the  bending  mo- 
ments at  those  supports  are  o,  —  ^w^>  —  ^>w^2,  and  o. 


Moments. 

Coefficients  of— 


ness. 

The  vertical  shears  at  the  supports  are  also  shown  in  the  fol- 
lowing table  for  any  number  of  spans  up  to  5.  The  space  repre- 
senting a  support  shows  in  its  left-hand  division  the  shear  on  the 
left  of  that  support  and  in  its  right-hand  division  the  shear  on  the 


9o 


RESTRAINED    BEAMS    AND    CONTINUOUS    BEAMS. 


IV. 


right  The  sum  of  the  two  shears  for  any  support  is,  of  course, 
the  reaction  of  that  support.  For  example,  in  a  beam  of  five  equal 
spans  the  reaction  at  the  second  support  is  ~pvl. 


Shears. 
Coefficients  of  wl. 


Naof 
spans. 


Fig:  35- 


It  will  be  seen  on  examination  that  the  numbers  in  any  oblique 
column  of  these  tables  follow  a  certain  law  of  increase  by  which 
it  is  possible  to  extend  them,  if  desired,  to  a  greater  number  of 
spans  than  are  here  given. 

As  an  example,  let  it  be  required  to  select  a  I  beam  to  span 

four  openings  of  8  feet  each,  the  load  per  span  being  500  pounds 

and  the  greatest  horizontal  stress  in  any  fiber  to  be  1 2  ooo  pounds 

per  square  inch.     The  required  beam  must  satisfy  formula  (4),  or, 

/_        M 

C      '     12  OOO 

where  M  is  the  maximum  moment.     From  the  table  it  is  seen 
that  the  greatest  negative  moment  is  that  at  the  second  support, 

or  _3-z£//2.     The  maximum  positive  moments  are, 
28 


For  first  span, 


V2 

max  M= —  ifl 

2w       's<* 

V* 


For  second  span,  max  M—  M2  + —  JL.  wl2. 

2iv       is68 

The  greatest  value  of  M  is  hence  at  the  second  support.     Then, 


ART.  50.     CONTINUOUS  BEAMS  WITH  UNEQUAL  SPANS.  91 

/  _  3  X  500  X  8  X  12 

c  12  ooo 

and  from  the  table  in  Art.  30  it  is  seen  that  a  light  /-inch  beam 
will  be  required. 

Prob.  82.  Find  what  I  beam  is  required  to  span  three  openings 
of  12  feet  each,  the  load  on  each  span  being  600  pounds,  and  the 
greatest  value  of  6"  to  be  12  ooo  pounds  per  square  inch. 

Prob.  83.  Draw  the  curve  of  moments  and  the  curve  of  shears 
for  the  case  of  three  equal,  spans  uniformly  loaded. 

ART.  50.    CONTINUOUS  BEAMS  WITH  UNEQUAL  SPANS. 

As  the  first  example,  consider  two  spans  whose  lengths  are  llt  1  2 
and  whose  loads  per  linear  unit  are  w1  and  w2.  The  theorem  of 
three  moments  in  (9)  then  reduces  to, 


and  hence  the  bending  moment  at  the  middle  support  is, 

M  =  _  ^.3  +  V.3 

8(/,  +  4) 

From  this  the  reaction  at  the  left  support  may  be  found  by  (8) 
and  the  bending  moment  at  any  point  by  (6). 

Next  consider  three  spans  whose  lengths  are  /I(  /2  and  /3,  loaded 
uniformly  with  zvlt  u>2,  w^.     The  bending  moments  at  the  second 
and  third  supports  are  M2  and  M  3.     Then  from  (9), 
0  +  M     =  -      w* 


and  the  solution  of  these  gives  the  values  of  M2  and  M^  A  very 
common  case  is  that  for  which  /2  =  /,  /T  =  /3=  nl,  and 
zvI  =  ^i>2  =  ze>3  =  iv.  For  this  case  the  solution  gives, 

M=W  =  -±+-?L       "^ 
2  -f  3»   '     4  ' 

Here  if  n  =  i  the  moments  become  —  —    as    shown    in    the 

10 
last  article. 


92        RESTRAINED  BEAMS  AND  CONTINUOUS  BEAMS.       IV. 

Whatever  be  the  lengths  of  the  spans  or  the  intensity  of  the 
loads,  the  theorem  of  three  moments  furnishes  the  means  of  find- 
ing the  bending  moments  at  the  supports.  Then  from  (8),  (7) 
and  (6)  the  vertical  shears  and  bending  moments  at  every  section 
may  be  computed.  Finally,  if  the  material  be  not  strained  beyond 
its  elastic  limit,  formula  (5)  may  be  used  to  determine  the  stiffness, 
while  (4)  investigates  the  strength  of  the  beam. 

Prob.  84.  A  heavy  1 2-inch  I  beam  of  36  feet  length  covers 
four  openings,  the  two  end  ones  being  each  8  feet  and  the  others 
each  10  feet  in  span.  Find  the  maximum  moment  in  the  beam. 
Then  determine  the  load  per  linear  foot  so  that  the  greatest  hori- 
zontal unit-stress  may  be  1 2  ooo  pounds  per  square  inch. 

Prob.  85.  A  continuous  beam  of  three  equal  spans  is  loaded 
only  in  the  middle  span.  Prove  that  the  reactions  of  the  end 

supports  due  to  this  load  are  —  — . 

20' 

ART.  51.     REMARKS  ON  THE  THEORY  OF  FLEXURE. 

The  theory  of  flexure  presented  in  this  and  the  preceding 
chapter  is  called  the  common  theory,  and  is  the  one  universally 
adopted  for  the  practical  investigation  of  beams.  It  should  not 
be  forgotten,  however,  that  the  axioms  and  laws  upon  which  it  is 
founded  are  only  approximate  and  not  of  an  exact  nature  like 
those  of  mathematics.  Laws  (A)  and  (B)  for  instance  are  true 
as  approximate  laws  of  experiment,  but  not  as  exact  laws  of 
science.  Law  (G)  indeed  rests  upon  so  slight  experimental  evi- 
dence that  it  is  more  of  a  hypothesis  than  an  established  truth. 
Objections  may  also  be  raised  against  the  validity  of  the  method 
of  resolving  the  internal  stresses  into  their  horizontal  and  vertical 
components,  and  against  the  formula  (3)  which  supposes  the  ver- 
tical shear  to  be  uniformly  distributed  over  the  cross-section. 

When  experiments  on  beams  are  carried  to  the  point  of  rup- 
ture and  the  longitudinal  unit-stress  5  computed  from  formula 
(4)  a  disagreement  of  that  value  with  those  found  by  direct  ex- 


ART.  51.          REMARKS    ON   THE   THEORY   OF    FLEXURE.  93 

periments  on  tension  or  compression  is  observed.  This  is  often 
regarded  as  an  objection  to  the  common  theory  of  flexure,  but  it 
is  in  reality  no  objection,  since  law  (G)  and  formula  (4)  are  only 
true  provided  the  elastic  limit  of  the  material  be  not  exceeded. 
Experiments  on  the  deflection  of  beams  furnish  on  the  other  hand 
the  most  satisfactory  confirmation  of  the  theory.  When  E  is 
known  by  tensile  or  compressive  tests  the  formulas  for  deflection 
are  found  to  give  values  closely  agreeing  with  those  observed. 
Indeed  so  reliable  are  these  formulas  that  it  is  not  uncommon  to 
use  them  for  the  purpose  of  computing  E  from  experiments  on 
beams.  If  however  the  elastic  limit  of  the  material  be  exceeded, 
the  computed  and  observed  deflections  fail  to  agree. 

On  the  whole  it  may  be  concluded  that  the  common  theory  of 
flexure  is  entirely  satisfactory  and  sufficient  for  the  investigation 
of  all  practical  questions  relating  to  the  strength  and  stiffness  of 
beams.  The  actual  distribution  of  the  internal  stresses  is  however 
a  matter  of  very  much  interest  and  this  will  be  discussed  in 
Chapter  VIII. 

The  theory  of  flexure  is  here  applied  to  continuous  beams  only 
for  the  case  of  uniform  loads.  It  should  be  said  however  that 
there  is  no  difficulty  in  extending  it  to  the  case  of  concentrated 
loads.  By  a  course  of  reasoning  similar  to  that  of  Art.  48  it 
may  be  shown  that  the  theorem  of  three  moments  for  single 
loads  is, 

Ml'  +  2M"(l'  +  I")  +  M"l"  =  —  Pl'*(k  —  #) 

-  P'l"*(2k  —  3/£2  -f  £*.) 

Here  as  in  Fig.  37  the  moments  at  three  consecutive  supports 
are  designated  by  M,  M"  and  M"'  and  the  lengths  of  the  two 
spans  by  /'  and  I".  P  is  any  load  on  the  first  span  at  a  distance 
kl'  from  the  left  support  and  P'  any  load  on  the  second  span  at 
a  distance  kl"  from  the  left  support,  k  being  any  fraction  less  than 
unity  and  not  necessarily  the  same  in  the  two  cases.  From  this 
theorem  the  negative  bending  moments  at  the  supports  for  any 
concentrated  loads  may  be  found,  and  the  beam  be  then  investi- 


94  RESTRAINED    BEAMS    AND    CONTINUOUS    BEAMS.  IV. 

gated  by  formulas  (6)  and  (4).  For  example,  if  a  beam  of  three 
equal  spans  be  loaded  with  P  at  the  middle  of  each  span  the  neg- 
ative moments  at  the  supports  are  each  Jz-Pl. 

20 

The  Journal  of  the  Franklin  Institute  for  March  and  April, 
1875,  contains  an  article  by  the  author  in  which  the  law  of  in- 
crease of  the  quantities  in  the  tables  of  Art.  45  is  explained  and 
demonstrated.  A  general  abbreviated  method  of  deducing  the 
moments  at  the  supports  for  both  uniform  and  concentrated  loads 
on  restrained  and  continuous  beams  is  given  in  the  Philosophical 
Magazine  for  September,  1875.  See  also  Van  Nostrand's  Science 
Series,  No.  25. 

•j^f"  Exercise  4.  Procure  six  sticks  of  ash  each  ^  X  ^  inches  and 
of  lengths  i,  2  and  3  feet.  Devise  and  conduct  experiments  to  test 
the  following  laws  :  First,  the  strength  of  a  beam  varies  directly 
as  its  breadth  and  directly  as  the  square  of  its  depth.  Second, 
the  stiffness  of  a  beam  is  directly  as  its  breadth  and  directly  as 
the  cube  of  its  depth.  Third,  a  beam  fixed  at  the  ends  is  twice 
as  strong  and  four  times  as  stiff  as  a  simple  beam  when  loaded  at 
the  middle.  Write  a  report  describing  and  discussing  the  ex- 
periments. 

Exercise  5.  Consult  Barlow's  Strength  of  Materials  (London, 
1837,)  and  write  an  essay  concerning  his  experiments  to  deter- 
mine the  laws  of  the  strength  and  stiffness  of  beams.  Consult 
also  Ball's  Experimental  Mechanics. 

Exercise  6.  In  order  to  test  the  theory  of  continuous  beams 
discuss  the  following  experiments  by  Francis  and  ascertain 
whether  or  not  the  ratio  of  the  two  observed  deflections  agrees 
with  theory.  "  A  frame  was  erected,  giving  4  bearings  in  the 
same  horizontal  plane,  4  feet  apart,  making  3  equal  spans,  each 
bearing  being  furnished  with  a  knife  edge  on  which  the  beam  was 
supported.  Immediately  over  the  bearings  and  secured  to  the 
same  frame  was  fixed  a  straight  edge,  from  which  the  deflections 
were  measured.  A  bar  of  common  English  refined  iron,  12  feet 
2^  inches  long,  mean  width  1.535  inches,  mean  depth  0.367 


ART.    51.          REMARKS    ON    THE   THEORY    OF    FLEXURE.  9$ 

inches,  was  laid  on  the  4  bearings,  and  loaded  at  the  center  of 
each  span  so  as  to  make  the  deflections  the  same,  the  weight  at 
the  middle  span  being  82.84  pounds  and  at  each  of  the  end  spans 
52.00  pounds.  The  deflections  with  these  weights  were, 

At  the  center  of  the  middle  span  0.281   inches. 

At  center  of  end  span,  0.275  and  0.284  inches, 

mean  0.280  inches. 

A  piece  3  feet  1 1  ^  inches  long  was  then  cut  from  each  end  of 
the  bar,  leaving  a  bar  4  feet  4^  inches  long,  which  was  replaced 
in  its  former  position  and  loaded  with  the  same  weight  (82.84 
pounds)  as  before,  when  its  deflection  was  found  to  be  1.059 
inches." 

Prob.  86.  A  beam  of  three  spans,  the  center  one  being  /  and 
the  side  ones  nl,  is  loaded  with  P  at  the  middle  of  each  span.  Find 
the  value  of  n  so  that  the  reactions  may  be  equal. 


COMPRESSION   OF   COLUMNS.  V. 


CHAPTER    V. 

ON  THE  COMPRESSION  OF  COLUMNS. 

ART.  52.      CROSS-SECTIONS  OF  COLUMNS. 

A  column  is  a  prism,  greater  in  length  than  about  ten  times 
its  least  diameter,  which  is  subject  to  compression.  If  the  prism 
be  only  about  four  or  six  times  as  long  as  its  least  diameter  the 
case  is  one  of  simple  compression,  the  constants  for  which  are 
given  in  Art.  6.  In  a  case  of  simple  compression  failure  occurs 
by  the  crushing  and  splintering  of  the  material,  or  by  shearing 
in  directions  oblique  to  the  length.  In  the  case  of  a  column, 
however,  failure  is  apt  to  occur  by  a  sidewise  bending  which  in- 
duces transverse  stresses  and  causes  the  material  to  be  highly 
strained  under  the  combined  compression  and  flexure. 

Wooden  columns  are  usually  square  or  round  and  they  may 
be  built  hollow.  Cast  iron  columns  are  usually  round  and  they 
are  often  cast  hollow.  Wrought  iron  columns  are  made  of  a 
great  variety  of  forms.  A  I  may  be  used  as  a  column,  but  they 
are  usually  made  of  three  or  more  different  shape-irons  riveted 
together.  The  Phoenix  column  is  made  by  riveting  together 
flanged  circular  segments  so  as  to  form  a  closed  cylinder.  It  is 
evident  that  a  square  or  round  section  is  preferable  to  an  unsym- 
metrical  one,  since  then  the  liability  to  bending  is  the  same  in  all 
directions.  For  a  rectangular  section  the  plane  of  flexure  will 
evidently  be  perpendicular  to  the  longer  side  of  the  cross-section, 
and  in  general  the  plane  of  flexure  will  be  perpendicular  to  that 
axis  of  the  cross-section  for  which  the  moment  of  inertia  is  the 


ART.  52.  CROSS-SECTIONS  OF  COLUMNS.  97 

least.  In  designing  a  column  it  is  hence  advisable  that  the  cross- 
section  should  be  so  arranged  that  the  moments  of  inertia  about 
the  two  principal  rectangular  axes  may  be  equal. 

For  instance,  let  it  be  required  to  construct  a  column  with  two 
I  shapes  and  two  plates  as  shown  in  Fig.  40.     The  I  beams  are 
to  be  light  lo-inch  ones  weighing  30  pounds 
per  linear  foot,  and  having  the  flanges  4.32 
inches  wide.     The  plates  are  to  be  ^  inch 
thick  and  it  is  required  to  find  their  length 
x  so  that  the  liability  to  bending  about  the 
two  axes  shown  in  the  figure  may  be  the 
same.     From  the  table  in  Fig.  30  it  is  ascer-       JJ^Jm\^myliiiil 
tained  that  the  moment  of  inertia  /  of  the  Fig^o. 

beam  about  an  axis  through  its  center  of 

gravity  and  perpendicular  to  the  web  is  150,  while  the  moment  of 
inertia  F  about  an  axis  through  the  same  point  and  parallel  to 
the  web  is  nearly  8.  Hence,  for  the  axes  shown  in  Fig.  40,  the 
moments  of  inertia  are, 

For  axis  perpendicular  to  plates, 

2°'5  *  *3  +2X8  +  2X9x(^  —  2.16) ' 

For  axis  parallel  to  plates, 
x  X  0.53 

2 -— ~    +   2   X   0.$X  X   5-2S2  +  2   X    150. 

Placing  these  two  expressions  equal,  the  value  of  x  is  found  to 
be  between  10^  and  n  inches. 

Prob.  87.  A  column  is  to  be  formed  of  two  light  1 2-inch  eye- 
beams  connected  by  a  lattice  bracing.  Find  the  proper  distance 
between  their  centers,  disregarding  the  moment  of  inertia  of 
the  latticing. 

Prob.  88.  Two  joists  each  2X4  inches  are  to  be  placed  6 
inches  apart  between  their  centers,  and  connected  by  two  others 
each  8  inches  wide  and  x  inches  thick  so  as  to  form  a  closed  hol- 
low rectangular  column.  Find  the  proper  value  of  x. 


gS  COMPRESSION  OF  COLUMNS.  V. 

ART.  53.     GENERAL  PRINCIPLES. 

If  a  short  prism  of  cross-section  A  be  loaded  with  the  weight 
P,  the  internal  stress  is  to  be  regarded  as  uniformly  distributed 

over  the  cross-section,  and  hence  the  compressive  unit-stress   Sc 

p 

is  — .     But  for  a  long  prism,  or  column,  this  is  not  the  case ; 
^jt 

p 

while  the  average  unit-stress  is  _,  the  stress  in  certain  parts  of 

A 

the  cross-section  may  be  greater  and  upon  others  less  than  this 
value  on  account  of  the  transverse  stresses  due  to  the  sidewise 
flexure.  Hence  in  designing  a  column  the  load  P  must  be  taken 
as  smaller  for  a  long  one  than  for  a  short  one,  since  evidently  the 
liability  to  bending  increases  with  the  length. 

Numerous  experiments  on  the  rupture  of  columns  have  shown 
that  the  load  causing  the  rupture  is  approximately  inversely  pro- 
portional to  the  length  of  the  column.  That  is  to  say,  if  there  be 
two  columns  of  the  same  material  and  cross-section  and  one 
twice  as  long  as  the  other,  the  long  one  will  rupture  under  about 
one-quarter  the  load  of  the  short  one. 

The  condition  of  the  ends  of  columns  exerts  a  great  influence 
upon  their  strength.     Class  (a)  includes  those  with  '  round  ends,' 
or  those  in  such  condition  that  they  are  free  to  turn  at  the  ends. 
Class  (c)  includes  those  whose  ends 
are  'fixed'  or  in  such  condition  that 
the  tangent  to  the  curve  at  the  ends 
always    remains   vertical.     Class   ($) 
includes   those   with  one  end  fixed 
and  the  other  round.    In  architecture 
it  is  rare  that  any  other  than  class  (c] 
is  used.     In  bridge  construction  and 
Fig.  41.  in    machines,    however,   columns    of 

classes  (G)  and  (a)  are  very  common. 
It  is  .evident  that  class  (c)  is  stronger  than  (b),  and  that  (ff)  is 


ART.  54. 


EULERS    FORMULA. 


99 


stronger  than  (c),  and  this  is  confirmed  by  all  experiments. 
Fig.  41  is  intended  as  a  symbolical  representation  of  the  three 
classes  of  columns,  and  not  as  showing  how  the  ends  are  ren- 
dered '  round '  and  '  fixed '  in  practical  constructions. 

The  theory  of  the  resistance  of  columns  has  not  yet  been  per- 
fected like  that  of  beams,  and  accordingly  the  formulas  for  prac- 
tical use  are  largely  of  an  empirical  character.  The  form  of  the 
formulas  however  is  generally  determined  from  certain  theoreti- 
cal considerations,  and  these  will  be  presented  in  the  following 
articles  as  a  basis  for  deducing  the  practical  rules. 

Prob.  89.  A  column  formed  of  two  I  beams  each  weighing  93 
pounds  per  yard  is  1 1  inches  square  and  3  feet  long.  What 
load  will  it  carry  with  a  factor  of  safety  of  5  ? 

ART.  54.     EULER'S  FORMULA. 

Consider  a  column  of  cross-section  A  loaded  with  a  weight  P 
under  whose  action  a  certain  small  sidewise  bend- 
ing occurs.  Let  the  column  be  round  or  free  to 
turn  at  both  ends  as  in  Fig.  42.  Take  the  origin 
at  the  upper  end,  and  let  x  be  the  vertical  and  y 
the  horizontal  co-ordinate  of  any  point  of  the  elas- 
tic curve.  The  general  equation  (4),  deduced  in 
Art  33,  applies  to  all  bodies  subject  to  flexure 
provided  the  bending  be  slight  and  the  elastic 
limit  of  the  material  be  not  exceeded.  For  the 
column  the  bending  moment  is  —  Pyt  and  hence, 


dx* 

The  first  integration  of  this -gives, 


But  when  y  =  the  maximum  deflection  J,  the  tangent   -Z.  =  o. 

dx 


I  GO  [      COMPRESSION    OF    COLUMNS. 

Hence  C  =  PA2,  and  by  inversion, 


V. 


The  second  integration  now  gives, 
x  =  (  —  j    arc  sin 

Here    C'  is  o  because  y  —  o  when  x  =  o.      Hence  finally  the 
equation  of  the  elastic  curve  of  the  column  is, 


y~.  +  C. 


j,  =      sn 
This  equation  is  that  of  a  sinusoid.  But  also  y  =  o  when  x  =  I. 

Hence  if  n  be  an  integer,  /(  —  )  '*  must  equal  UK,  or, 

772.7.2 
P  =  El—, 

/2 

which  is  Euler's  formula  for  the  resistance  of  columns.     This  re- 
duces the  equation  of  the  sinusoid  to, 


The  three 'curves  for  n=  I,  n  =  2,  and  n  =t  3  are  shown  in  Fig. 

43.     In  the  first  case  the  curve 

n  n  f! 

is  entirely  on  one  side  of  the 
axis  of  x,  in  the  second  case  it 
crosses  that  axis  at  the  middle, 
and  in  the  third  case  it  crosses 
at  ^/and  ^/,  the  points  of  cross- 
ing being  also  inflection  points 


n=2 
Fig".  43. 


71=3 


where  the  bending  moment  is 
zero.      Evidently   the    greatest 


deflection  will  occur  for  the  case  where  n  =  i,  and  this  is  the 
most  dangerous  case.     Hence, 

(a)  P  = 


is  Euler's  formula  for  columns  with  round  ends. 


ART.  54.  EULER'S  FORMULA.  101 

A  column  with  one  end  fixed  and  the  other  round  is  closely 
represented  by  the  portion  b'b"  of  the  second  case,  b'  being  the 
fixed  end  where  the  tangent  to  the  curve  is  vertical.  Here  n  =  2, 
and  the  length  b'b"  is  three-fourths  of  the  entire  length,  hence, 

(b)  P  =  f^- 

is  Euler's  formula  for  columns  with  one  end  fixed  and  the  other 
round. 

A  column  with  fixed  ends  is  represented  by  the  portion  c'c"  of 
the  case  c.  Here  n  =  3,  and  the  length  c'c"  is  three-fourths  of 
the  entire  length,  hence, 


is  Euler's  formula  for  columns  with  fixed  ends. 

From  this  investigation  it  appears  that  the  relative  resistances 
of  three  columns  of  the  classes  (a),  (b)  and  (<:)  are  as  the  numbers 
1,2^  and  4,  when  the  lengths  are  the  same,  and  this  conclusion 
is  approximately  verified  by  experiments.  It  also  appears  that, 
if  the  resistance  of  three  columns  of  the  classes  (a),  (H)  and  (c)  are 
to  be  equal,  their  lengths  must  be  as  the  numbers  I,  \y2  and  2. 

The  moment  of  inertia  /  in  the  above  formulas  is  taken  about 
a  neutral  axis  of  the  cross-section  perpendicular  to  the  plane  of 
the  flexure,  and  in  general  is  the  least  moment  of  inertia  of  that 
cross-section,  since  the  column  will  bend  in  the  direction  which 
offers  the  least  resistance.  For  a  rectangular  column  whose 
greatest  side  is  b  and  least  side  d,  the  formulas  may  be  written, 


„  i  / 

P  =  ---  ,         where  m  =  i,  2%,  or  4. 

For  a  cylindrical  column  of  diameter  d  the  formulas  are, 

„        mx^Ed*  / 

P  =   ----  ,        where  m  =  1  ,  2  %  ,  or  4. 
64/2 

Hence  the  strength  of  a  column  varies  directly  as  its  cross-section 
and  directly  as  the  square  of  its  least  diameter  or  side.  In  genera] 


IO2  COMPRESSION   OF   COLUMNS.  V. 

if  r  be  the  least  radius  of  gyration  of  the  cross-section  the  value 
of  /  is  Ar*  and  the  formula  may  be  written, 

P_=  mr:^         ^^  m  =  ^  ^  Qr  ^ 

which  shows  that  P  varies  as  the  square  of  the  ratio  of  r  to  /. 

The  maximum  deflection  J  is  indeterminate,  so  that  the  load 
P,  given  by  Euler's  formula,  is  merely  the  load  which  causes  the 
column  to  bend.  Practically  the  bending  of  a  column  is  the 
beginning  of  its  failure. 

Prob.  90.  Show  that  Euler's  formula  for  the  case  of  a  column 

P  7T2J£?'2 

fixed  at  one  end  and  entirely  free  at  the  other  is  —  —  -  --- 

A  4/2 

ART.  55.     HODGKINSON'S  FORMULAS. 

Euler's  formula  gives  valuable  information  regarding  the  laws 
of  flexure  of  columns,  but  is  difficult  of  direct  practical  application 
because  it  indicates  no  relation  between  the  load  P  and  the  greatest 
internal  compressive  unit-stress.  It  shows  that  the  strength  of 
cylindrical  columns  varies  directly  as  the  fourth  power  of  the  diam- 
eter and  inversely  as  the  square  of  the  length.  Hodgkinson  in  his 
experiments  observed  that  this  was  approximately  but  not  exactly 
the  case.  He  therefore  wrote  for  each  kind  of  columns  the  ana- 
logous expression, 


and  determined  the  constants  a,  ft  and  d  from  the  results    of  his 
experiments,  thus  producing  empirical  formulas. 

Let  P  be  the  crushing  load  in  gross  tons,  d  the  diameter  of  the 
column  in  inches,  and  /  its  length  in  feet.  Then  Hodgkinson's 
empirical  formulas  are, 

For  solid  cast  iron  cylindrical  columns, 

P=  14.9  -  for  round  'ends, 

P  =  44.2  —  —  for  flat  endsr 

IT-  -03 


ART.  56.  TREDGOLD'S  FORMULA.  103 

For  solid  wrought  iron  cylindrical  columns, 

dz'lf> 
P  =  42 for  round  ends, 

P  =  134 for  flat  ends. 

These  formulas  indicate  that  the  ultimate  strength  of  flat-ended 
columns  is  about  three  times  that  of  round-ended  ones.  The 
experiments  also  showed  that  the  strength  of  a  column  with  one 
end  flat  and  the  other  end  round  is  about  twice  that  of  one  having 
both  ends  round.  Hodgkinson's  tests  were  made  upon  small  col- 
umns and  his  formulas  are  not  so  reliable  as  those  which  will  be 
given  in  the  following  articles.  For  small  cast  iron  columns  how- 
ever the  formulas  are  still  valuable. 

By  the  help  of  logarithms  it  is  easy  to  apply  these  formulas  to 
the  discussion  of  given  cases.  Usually  P  will  be  given  and  d  re- 
quired, or  d  be  given  and  P  required.  By  using  assumed  factors 
of  safety  the  proper  size  of  cylindrical  columns  to  carry  given  loads 
may  also  be  determined.  These  formulas,  it  should  be  remem- 
bered, do  not  apply  to  columns  shorter  than  about  thirty  times 
their  least  diameters.  The  word  flat  used  in  this  article  is  to  be 
regarded  as  equivalent  to  fixed. 

Prob.  91.  A  cast  iron  cylindrical  column  with  flat  ends  is  3 
inches  diameter  and  8  feet  long.  What  load  will  cause  it  to  fail  ? 

Prob.  92.  A  cast  iron  cylindrical  column  with  flat  ends  is  to  be 
7  feet  long  and  carry  a  load  of  200  ooo  pounds  with  a  factor  of 
safety  of  6.  Find  the  proper  diameter. 

ART.  56.      TREDGOLD'S   FORMULA. 

The  formulas  of  Euler  are  defective  because  they  contain  no 
constant  indicating  the  working  or  ultimate  compressive  strength 
of  the  material  and  because  they  apply  only  to  long  columns. 
Hodgkinson's  formulas  are  unsatisfactory  for  similar  reasons  and 
because  they  do  not  well  represent  the  results  of  later  experiments; 


104 


COMPRESSION    OF   COLUMNS. 


V. 


Tredgold's  formula  was  deduced  to  remedy  these  defects.  It  may 
be  established  by  the  following  considerations. 

Let  the  column  be  of  rectangular  section,  the  area  being  A,  the 
least  side  d,  the  greatest  side  b,  and  the  length  /.  Let  P  be  the 
load  upon  it.  The  average  compressive  unit- 

p 
stress  at  any  section  is  — ,   but    in      conse- 

^± 

quence  of  the  sidewise  deflection  this  is  in- 
creased on  the  concave  side  and  decreased 
on  the  convex  side  by  an  amount  .S.  From 
the  fundamental  equation  (4)  the  value  of  5 

is ,     and  if  J  be  the  maximum  deflection 


bd* 

the  greatest  value  of  6"  is 


6PJ 


Now  if  5  he 


the  total  compressive  unit-stress  on  the  con- 

p 
cave  side  S  =  Sc  —  —,  and  hence, 

A 


S  — 


6PJ      6P4 


Accordingly  the  value  of  Sc  is 

~  _  P       P  64 
c~~A+~A~d' 

The  value  of  J  is  unknown,  but  if  the  curve  of  deflection  be 
an  arc  of  a  circle,  which  it  is  very  nearly,  J  equals  approximately 

I2 
— ,  in  which  R  represents  the  radius  of  curvature  of  the  column. 

Now,  as  in  Art.  33,  the  value  of  R  for  the  same  unit-stress  J> 
varies  directly  as  d  and  inversely  as  E.  Hence  J  may  be  taken 
as  varying  directly  as  /2  and  inversely  as  d.  Accordingly  if  k  be 
a  number  depending  upon  the  kind  of  material  and  the  arrange- 
ment of  the  ends  of  the  column,  the  value  of  Sf  may  be  written, 


4 


-. 
4<C 


ART.  56.  TREDGOLD'S  FORMULA.  105 

p 

From  this  the  value  of  the  unit  load  is, 

A 


which  is  Tredgold's  formula  for  resistance  of  columns. 

The  quantity  k  can  not  be  determined  theoretically.  As  the 
above  reasoning  shows,  its  value  varies  with  the  form  of  cross- 
section  as  well  as  with  the  kind  of  material  and  the  arrangement 
of  the  ends  of  the  column.  For  instance,  the  value  of  k  is  not 
the  same  for  a  circular  section  with  diameter  d  as  for  a  rectangu- 
lar section  whose  least  side  is  d.  It  is  however  not  uncommon 
to  find  this  formula  stated  as  applicable  to  any  cross-section  whose 
least  diameter  is  d. 

In  order  to  determine  k  recourse  must  be  had  to  experiments. 
These  are  usually  conducted  by  loading  columns  to  the  point  of 
rupture.  P,  A,  I  and  d  are  known  and  thus  the  constants  Sc  and 
k  may  be  computed.  Theoretically  Sc  is  the  ultimate  compressive 
strength  of  the  material  and  the  values  found  for  it  by  experi- 
ments on  columns  agree  roughly  with  those  deduced  by  the  direct 
crushing  of  short  specimens.  The  value  of  k  is  always  less  than 
unity  and  it  is  subject  to  great  variation,  even  in  columns  of  the 
same  material.  For  a  column  with  round  ends  k  was  regarded 
by  Tredgold  and  Gordon  as  being  four  times  as  great  as  for  a  col- 
umn with  fixed  ends,  since  both  experiment  and  theory  indicate 
that  a  fixed-ended  column  of  length  2.1  has  the  same  strength  as  a 
round-ended  column  of  length  /.  Therefore  for  the  ultimate 
strength  of  columns, 

P          S 

For  fixed  ends,  —  = £ — , 

P          S 
For  round  ends,  -  = c- 


106  COMPRESSION    OF    COLUMNS.  V. 

The  following  values  of  Sc  and  k  were  deduced  by  Gordon  from 

Hodgkinson's  experiments,  and  are  given  by  Rankine, 

For  stone  and  brick,  5  =  see  Art.  6,  k  =  -^-> 

c  OOO 

For  timber  (rectangular  sections),  Sc  =    7  200,       k  =  — ' 

For  cast  iron  cylinders,  Sc  =  80  ooo,       k  —  -^-, 

For  wrought  iron  (rectangular  sections),  Sc  =  36  ooo,       k  =  — — 

These  values  of  Sc  are  in  pounds  per  square  inch,  while  those  of 
k  are  abstract  numbers. 

Tredgold's  formula  is  sometimes  used  under  the  name  of  Gor- 
don's formula,  The  reasoning  by  which  it  is  deduced  is  not  en- 
tirely satisfactory  and  it  often  fails  to  properly  represent  the  results 
of  experiment. 

Prob.  93.  Find  the  values  of  Sc  and  k  from  the  two  following 
experiments  on  flat-ended  Phoenix  columns.  The  sectional  area 
of  each  column  was  1 2  square  inches  and  the  exterior  diameter  8 
inches.  The  length  of  the  first  column  was  25  feet  and  it  failed 
under  a  load  of  420  ooo  pounds.  The  length  of  the  second  col- 
umn was  10  feet  and  it  failed  under  a  load  of  478  ooo  pounds. 

ART.  57.     GORDON'S  FORMULA. 

The  formula  which  seems  to  most  satisfactorily  represent  the 
results  of  experiments  will  now  be  deduced.  It  is  often  called 
Gordon's  formula  and  sometimes  Rankine's  formula,  and  occasion- 
ally it  is  referred  to  as  "  Gordon's  formula  modified  by  Rankine." 
It  does  not  appear  however  that  either  Gordon  or  Rankine  devel- 
oped it  in  its  general  form  or  used  it  for  the  discussion  of  experi- 
ments. The  name  of  Gordon  will  here  be  applied  to  it  because 
that  is  most  frequently  used  and  because  of  the  lack  of  a  better 
appellation.  It  is  similar  to  Tredgold's  formula,  but  has  the  ad- 
vantage of  being  applicable  to  any  form  of  cross-section. 

Let  P  be  the  load  on  the  column,  /  its  length,  A  the  area  of  its 
gross-section,  /  the  moment  of  inertia  and  r  the  radius  of  gyra- 


ART.  57. 


GORDON  S    FORMULA. 


ID/ 


tion  of  that  cross-section  with  reference  to  a  neutral  axis  perpen- 
dicular to  the  plane  of  flexure,  and  c  the  shortest  distance  from 
that  axis  to  the  remotest  fiber  on  the  concave  side.  The  average 

compressive  unit-stress  on  any  cross-sec- 

p 
tion  is  —  but  in  consequence  of  the  flex- 

A 

ure  this  is  increased  on  the  concave  side 

and  decreased  on  the  convex  side.    Thus 

p 
in  Fig.  45  the  average  unit-stress  is 

A 

represented  by  ab,  but  on  the  concave  side 

this  is  increased  to  eg  and  on  the  convex 

side  decreased  to  eh.      The  triangles  bdg  and  bfh  represent  the 

effect  of  the  flexure  exactly  as  in  the  case  of  beams,  dg  indicating 

the  greatest  compressive  and  hf  the  greatest  tensile  unit-stress  due 

to  the  bending.     Let  the  total  maximum  unit-stress  be  denoted 

by  J>  and  the  part  due  to  the  flexure  be  denoted  by  6".     Then, 


Me 
Now,  from  the  fundamental  formula  (4),  theflexural  stress  is   , 

where  M  is  the  external  bending  moment,  which  for  a  column  has 
its  greatest  value  when  M—  PJ,  A  being  the  maximum  deflection. 
/  —  Ar2  is  the  well  known  relation  between  /  and  r.  Hence  the 
value  of  5  is, 

s_**.«f. 


By  analogy  with  the  theory  of  beams,  as  in  Art.  37,  the  value  of 

/2 

J  may  be  regarded  as  varying  directly  as    _.      Hence   if  q  be  a 

c 

quantity  depending  upon  the  kind  of  material  and  the   condition 
of  the  ends,  the  total  unit-stress  is, 


IO8  COMPRESSION    OF   COLUMNS. 

This  may  now  be  written  in  the  usual  form, 
0°)  -*  =  ~^. 


which  is  Gordon's  formula  for  the  investigation  of  columns. 

The  above  reasoning  has  been  without  reference  to  the  arrange- 
ment of  the  ends  of  the  column.  By  Art.  54  it  is  known  that  a 
column  with  round  ends  must  be  one-half  the  length  of  one  with 
fixed  ends  in  order  to  be  of  equal  strength,  and  that  a  column 
with  one  end  fixed  and  the  other  round  must  be  three-fourths  the 
length  of  one  with  fixed  ends  in  order  to  be  of  equal  strength. 
Therefore  if  q  be  the  constant  for  fixed  ends,  (^fq  will  be  the 
constant  for  one  end  fixed  and  the  other  round,  and  2zq  will  be 
the  constant  for  both  ends  round. 

The  values  of  q  to  be  taken  for  use  in  formula  (10)  for  the  ex- 
amples and  problems  of  this  chapter  may  be  the  following  rough 
values,  unless  otherwise  stated,  while  the  values  of  the  ultimate 
compressive  unit-stress  Sc  will  be  taken  from  Art.  6. 


Material. 

Both  Ends 
Fixed. 

Fixed 
and  Round. 

Both  Ends 
Round. 

Timber 
Cast  Iron 
Wrought  Iron 
Steel 

I 

I.;8 

4 

3  coo 
i 

3  ooo 
1.78 

3  ooo 
4 

5  ooo 

5  ooo 
1.78 

5  ooo 
4 

36  ooo 
i 

36  ooo 

1.78 

36  ooo 
4 

25  ooo 

25  ooo 

25  ooo 

The  very  wide  variation  in  the  values  of  q  found  from  different 
experiments  shows  however  that  little  dependence  can  be  placed 
upon  average  results.  In  any  practical  case  of  importance  an 


ART.    58.          RADIUS    OF    GYRATION   OF   CROSS-SECTION.  IO$ 

effort  should  be  made  to  ascertain  values  of  Sc  and  q  for  the  special 
kind  of  columns  on  hand. 

Prob.  94.  Plot  the  curve  represented  by  formula  (10)  for  cases 
of  wrought  iron  columns  with  fixed  and  with  round  ends,  taking 

P  I 

the  values  of  —  as  ordinates  and  the  values  of  —  as  abscissas. 
A  r 

ART.  58.     RADIUS  OF  GYRATION  OF  CROSS-SECTIONS. 

The  radius  of  gyration  of  a  surface  with  reference  to  an  axis  is 
equal  to  the  square  root  of  the  ratio  pf  the  moment  of  inertia  of 
the  surface  referred  to  the  same  axis  to  the  area  of  the  figure.  Or 
if  r  be  radius  of  gyration,  /  the  moment  of  inertia  and  A  the  area 
of  the  surface,  then  /  =  Ar2. 

In  the  investigation  of  columns  by  formula  (10)  the  value  of  r2 
is  required,  r  being  the  least  radius  of  gyration.  These  values 
are  readily  derived  from  the  expressions  for  the  moment  of  inertia 
given  in  Art.  22,  the  most  common  cases  being  the  following. 

For  a  rectangle  whose  least  side  is  d.          r2  =  — 

12 

For  a  circle  of  diameter  d,  r2  =  — 

16 

For  a  triangle  whose  least  altitude  is  d,         r2  =  --. 

Io 

u  n  d*  +  d'* 

For  a  hollow  square  section,  r2-  =  —      — . 

12 

For  a  hollow  circular  section,  r2  = . 

16 

For  I  beams  and  other  shapes  r2  is  found  by  dividing  the  least 
moment  of  inertia  of  the  cross-section  by  the  area  of  that  cross- 
section.  For  instance,  by  the  help  of  the  table  in  Art.  30,  the 
least  value  of  r2  for  a  light  12-inch  eye  beam  is  found  to  be 
~6-  =  0.87  inches2. 

Prob.  95.  An  angle  iron  is  3  X  4  X  0.5  inches.  Find  the  least 
moment  of  inertia  and  least  radius  of  gyration. 


IIO  COMPRESSION  OF  COLUMNS.  V. 

ART.  59.     INVESTIGATION  OF  COLUMNS. 

The  investigation  of  a  column  consists  in  determining  the  max- 
imum compressive  unit-stress  Sc  from  formula  (10).  The  values 
of  P,  A,  I  and  r  will  be  known  from  the  data  of  the  given  case 
and  q  is  known  from  the  results  of  previous  experiments.  Then, 


and,  by  comparing  the  computed  value  of  Sc  with  the  ultimate 
strength  and  elastic  limit  of  the  material,  the  factor  of  safety  and 
the  degree  of  stability  of  th:  column  may  be  inferred. 

For  example,  consider  a  hollow  cast  iron  column  of  rectangu- 
lar section,  the  outside  dimensions  being  4X5  inches  and  the 
inside  dimensions  3X4  inches.  Let  the  length  be  18  feet,  the 
ends  fixed,  and  the  load  be  80  ooo  pounds.  Here  P  =  80  ooo, 

A  =8  square  inches,  /  =  216  inches.  From  the  table  q  =  _  L_ 

3000' 
From  Art.  58, 

_   5  X  43  -  4  X  32  _ 

12  X  8 
Then  the  substitution  of  these  values  gives, 

80  ooo/      .    216  X  216  \ 
Se  =  —  —    i  +  -  =  42  ooo  pounds  per  square  in. 


Here  the  average  unit-ctress  is  10  ooo  pounds  per  square  inch, 
but  the  flexure  has  increased  that  stress  on  the  concave  side  to 
42  ooo  pounds  per  square  inch  so  that  the  factor  of  safety  is  only 
about  two. 

Prob.  96.  A  cylindrical  wrought  iron  column  with  fixed  ends  is 
12  feet  long,  6.36  inches  in  exterior  diameter,  6.02  inches  in  in- 
terior diameter,  and  carries  a  load  of  98  ooo  pounds.  Find  its 
factor  of  safety. 

Prob.  97.  A  pine  stick  3X3  inches  and  12  feet  long  is  used 
as  a  column  with  fixed  ends.  Find  its  factor  of  safety  under  a 
load  of  3  ooo  pounds.  If  the  length  be  only  one  foot,  what  is 
the  factor  of  safety. 


ART.    6l.  DESIGNING    OF   COLUMNS.  Ill 

ART.  60.     SAFE  LOADS  FOR  COLUMNS. 

To  determine  the  safe  load  for  a  given  column  it  is  necessary 
to  first  assume  the  allowable  working  unit-stress  Sc.  Then  from 
formula  (10)  the  safe  load  is, 

P         S<A 


Here  A,  I  and  r  are  known  from  the  data  of  the  given  problem 
and  q  is  taken  from  the  table  in  Art.  58. 

For  example,  let  it  be  required  to  determine  the  safe  load  for  a 
fixed-ended  timber  column,  3X3  inches  square  and  1  2  feet  long, 
so  that  the  greatest  compressive  unit-stress  may  be  800  pounds 
per  square  inch.  From  the  formula,  / 

800  X  Q 
P  =  ---  —  -  =  about  700  pounds. 

3  ooo  X~3~2 

A  short  prism  3X3  inches  should  safely  carry  ten  times  this 
load. 

Prob.  98.  Find  the  safe  load  for  a  heavy  wrought  iron  I  of  1  5 
inches  depth  and  for  lengths  of  5,  loand  15  feet  when  used  as 
columns  with  fixed  ends. 

ART.  61.      DESIGNING  OF  COLUMNS. 

When  a  column  is  to  be  selected  or  designed  the  load  to  be 
borne  will  be  known,  as  also  its  length  and  the  condition  of  the 
ends.  A  proper  allowable  unit-stress  Sc  is  assumed,  suitable  for 
the  given  material  under  the  conditions  in  which  it  is  used.  Then 

from  formula  (i)  the  cross-section  of  a  short  column  or  prism  is 

p 

—  and  it  is  certain  that  a  greater  value  of  the  cross-section  than 

this  will  be  required.  Next  assume  a  form  and  area  A,  find  r2, 
and  from  the  formula  (  i  o)  compute  Sc.  If  the  computed  value 
agrees  with  the  assumed  value  the  correct  size  has  been  selected. 


112  COMPRESSION    OF    COLUMNS.  V. 

If  not,  assume  a  new  area  and  compute  Sc  again,  and  continue 
the  process  until  a  proper  agreement  is  attained. 

For  example,  a  hollow  cast  iron  rectangular  column  of  18  feet 
length  is  to  carry  a  load  of  60  ooo  pounds.  Let  the  working 
strength  Sc  be  15  ooo  pounds  per  square  inch.  Then  for  a  short 
length  the  area  required  would  be  four  square  inches.  Assume 
then  that  about  6  square  inches  will  be  needed.  Let  the  section 
be  square,  the  exterior  dimensions  6X6  inches,  and  the  interior 
dimensions  5^  X  $^2  inches.  Then  ^  =  5.75,  /=  18  X  12, 

^=60000,   g  = -—-,    r2  =  5.52,  and  from  (10), 
5  ooo 

60  ooo/  i82  X  I22     \ 

S.  =  —        •  I  i  H -  I  =  about  30  ooo, 

5.75   V        5  ooo  X  5.52  / 

which  shows  that  the  dimensions  are  much  too  small. 

Again  assume  the  exterior  side  as  6  inches  and  the  interior  as 
5  inches.  Then  A  =  II,  r2  =  5.08,  and 

c        60  ooo/  i82  X  I22     \ 

S,  = —I  i  + -  I  =  about  15  700. 

ii     V          5  ooo  X  5.08  / 

As  this  is  very  near  the  required  working  stress  it  appears  that 
these  dimensions  very  nearly  satisfy  the  imposed  conditions. 

In  many  instances  it  is  possible  to  assume  all  the  dimensions 
of  the  column  except  one  and  then  after  expressing  A  and  r  in 
terms  of  this  unknown  quantity  to  introduce  them  into  (10)  and 
solve  the  problem  by  finding  the  root  of  the  equation  thus  formed. 
For  example  let  it  be  required  to  find  the  size  of  a  square  wooden 
column  with  fixed  ends  and  24  feet  long  to  sustain  a  load  of 
100  ooo  pounds  with  a  factor  of  safety  of  10.  Here  let  x  be  the 

unknown  side;  then  A  =x*  and  r2  =  —      From  (10), 

12 


x*  3000X*8 

By  reduction  this  becomes, 

8;t4  —  i  ooox2  =  25 1  776, 
the  solution  of  which  gives  14.6  inches  for  the  side  of  the  column. 


ART.  62.          EXPERIMENTS  ON  COLUMNS.  113 

Prob.  99.  A  hollow  cylindrical  cast  iron  column  is  to  be  de- 
signed to  carry  a  load  of  200  ooo  pounds.  Its  length  is  to  be  12 
feet,  its  ends  flat  or  fixed,  its  exterior  diameter  6  inches  and  the 
allowable  unit-stress  1 5  ooo  pounds  per  square  inch.  Find  the 
proper  interior  diameter. 

Prob.  100.  Find  the  size  of  a  square  wooden  column  with  fixed 
ends  and  12  feet  in  length  to  sustain  a  load  of  100  ooo  pounds 
with  a  factor  of  safety  of  10.  Find  also  its  size  for  round  ends. 

ART.  62.      EXPERIMENTS  ON  COLUMNS. 

It  is  impossible  to  present  here  even  a  summary  of  the  many 
experiments  that  have  been  made  to  determine  the  laws  of  re- 
sistance of  columns.  The  interesting  tests  made  by  Christie  in 
1883  for  the  Pencoyd  Iron  Works  will  however  be  briefly  de- 
scribed on  account  of  their  great  value  and  completeness  as  re- 
gards wrought  iron  struts,  embracing  angle,  tee,  beam  and  chan- 
nel sections.  See  Transactions  of  the  American  Society  of  Civil 
Engineers,  April,  1884. 

The  ends  of  the  struts  were  arranged  in  different  methods ; 
first  flat  ends  between  parallel  plates  to  which  the  specimen  was 
in  no  way  connected ;  second,  fixed  ends,  or  ends  rigidly  clamped ; 
third,  hinged  ends,  or  ends  fitted  to  hemispherical  balls  and 
sockets  or  cylindrical  pins;  fourth,  round  ends,  or  ends  fitted 
to  balls  resting  on  flat  plates. 

The  number  of  experiments  was  about  300,  of  which  about 
one-third  were  upon  angles,  and  one-third  upon  tees.  The  qual- 
ity of  the  wrought  iron  was  about  as  follows  :  elastic  limit  32  ooo 
pounds  per  square  inch,  ultimate  tensile  strength  49  600  pounds 
per  square  inch,  ultimate  elongation  18  percent  in  8  inches.  The 
length  of  the  specimens  varied  from  6  inches  up  to  16  feet,  and 
the  ratio  of  length  to  least  radius  of  gyration  varied  from  20  to 
480.  Each  specimen  was  placed  in  a  Fairbanks'  testing  machine 
of  50  ooo  pounds  capacity  and  the  power  applied  by  hand  through 
a  system  of  gearing  to  two  rigidly  parallel  plates  between  which 


114 


COMPRESSION   OF   COLUMNS. 


V. 


the  specimen  was  placed  in  a  vertical  position.  The  pressure  or 
load  was  measured  on  an  ordinary  scale  beam,  pivoted  on  knife 
edges  and  carrying  a  moving  weight  which  registered  the  pressure 
automatically.  At  each  increment  of  5  ooo  pounds,  the  lateral 
deflection  of  the  column  was  measured.  The  load  was  increased 
until  failure  occured. 

The  following  are  the  combined  average  results  of  these  care- 
fully conducted  experiments.     The  first  column  gives  the  values 


of ,  and   the   other   columns 

r 


the 


value  of  —  or  the  ultimate 
A 


load  per  inch  of  cross-section.    From  these  results  it  will  be  seen 


Length  divided 
by  Least  Rad. 
of  Gyration. 

Flat  Ends. 

Fixed  Ends. 

Hinged  Ends. 

Round  Ends. 

20 

46  ooo 

46  ooo 

46  ooo 

44  ooo 

40 

40  ooo 

40  ooo 

40  ooo 

36  500 

60 

36  ooo 

36  ooo 

36  ooo 

30500 

80 

32  ooo 

32  ooo 

31  500 

25  ooo 

100 

29  800 

30  ooo 

28  ooo 

20  500 

120 

26  300 

28  ooo 

24  300       1  6  500 

140 

23  500 

25  500 

21  OOO 

12  8OO 

160 

20  ooo 

23000 

16  500 

9  5°o 

1  80 

16800 

20  ooo 

12  800 

7  50°  . 

2OO 

14  500 

17  500 

10  800 

6  ooo 

220 

12  700 

15  ooo 

8800 

5  ooo 

240 

II  2OO 

13000 

7  5°o 

4300 

260 

9800 

II  OOO 

6  500 

3  800 

280 

8  500 

IO  OOO 

5  700 

3  200 

300 

7  200 

9  ooo 

5  ooo 

2  800 

320 

6000 

8  ooo 

4  5°°" 

2  500 

340 

5  loo 

7  ooo 

4000 

2  IOO 

360 

4300 

6  500 

3  5°o 

I  000 

380 

3  BOO 

5  800 

3000 

I  7OO 

400 

3  ooo 

5  200 

2  500 

I  500 

420 

2  500 

4800 

2  300 

I  300 

440 

2  200 

4300 

2  IOO 

460 

2  OOO 

3  800 

I  900 

480 

I  900 

I  800 

ART.  63.  THE  THEORY  OF  COLUMNS.  115 

that  when  the  strut  is  short  there  is  no  practical  difference  in  the 
strength  of  the  four  classes,  and  that  when  the  strut  is  long  there 
is  but  little  difference  between  those  with  flat  and  hinged  ends. 
The  strength  of  the  long  columns  with  fixed  ends  appears  to  be 
about  3^  times  that  of  the  round-ended  ones. 

Prob.  101.  Plot  the  above  experiments,  taking  the  values  of 

/  P 

-  as  abscissas  and  those  of  —  as  ordinates. 

r  A 

Prob.  1 02.  What  load  will  cause  the  rupture  of  a  wrought  iron 
strut  of  an  angle  section  i  X  I  X  #J  inch  and  5  feet  long  when 
acting  with  flat  ends  ?  Ans.  About  7  200  pounds. 

ART.  63.     ON  THE  THEORY  OF  COLUMNS. 

It  has  been  already  remarked  that  the  theory  of  columns  is  in 
a  very  incomplete  condition  compared  with  that  of  beams.  A 
satisfactory  formula  for  the  resistance  of  columns  should  be  of 
such  a  nature  that  for  a  short  block  which  fails  by  pure  crushing 
it  would  reduce  to  the  equation  P  =  ASC,  while  for  a  long  strut 
which  fails  by  bending  it  would  reduce  to  an  expression  like 
Euler's.  The  formula  of  Gordon  conforms  partly  to  this  require- 
ment, but  the  fact  that  it  is  impossible  to  determine  values  of  q  of 
general  applicability  indicates  that  q  is  not  a  constant,  and  that 
the  reasoning  by  which  it  is  deduced  is  faulty.  Nevertheless 
Gordon's  formula  applies  so  well  to  columns  of  medium  length 
that  it  is  extensively  employed  in  this  country  in  the  manner 
illustrated  in  the  preceding  articles. 

For  long  columns  Euler's  formula  often  represents  fairly  the 
results  of  experiments,  and  since  it  contains  /  it  may  be  adapted 
to  any  form  of  cross-section.  Thus  /  =  Ar2,  and, 

For  round  ends, 

A 

p 

For  fixed  ends, 

A 


Il6  COMPRESSION    OF    COLUMNS.  V. 

For  wrought  iron  E  equals  about  25  ooo  ooo  pounds  per  square 
inch  and  hence  for  round  ends, 

if      -  =      200,  300,  400, 

r 

p 

—  =  6  250,       2  800,       i  600, 

A 

and  these  agree  well  with  the  experimental  values  given  in  the 
last  article. 

In  conclusion  it  may  be  well  to  show  that  Gordon's  formula 
when  properly  deduced  is  essentially  of  the  same  form  as  Euler's, 
and  that  the  number  q  cannot  be  a  true  constant.  For  this  pur- 
pose consider  the  reasoning  of  Art.  57,  and  as  there  take  the  total 
compressive  unit-stress  Sc  on  the  concave  side  as  equal  to  the 
sum  of  the  average  unit-stress  and  the  flexural  unit-stress,  or, 


From  the  fundamental  formula  (4)  the  value  of  .S"  is, 
z_PAc 
"A?' 

Now  to  express  J  in  terms  of  /,  consider  the  case  of  columns 
with   round   ends   which   deflect   into  a  sinusoid  curve,  whose 

equation  according  to  Art.  54  is, 

.     .     KX 

y  =  A  sin 

The  second  derivative  of  y  with  respect  to  x  is, 

d?y        Jn*    .     KX 

— —  = sin 

dx2         I2  I 

For   the  middle  of  the  column  where  x  =  *£l,  the  curvature 
hence  is, 


But  the  investigation  of  Art.  33  shows  also  that  SR  =  EC.  Hence, 


ART.    63.  THE   THEORY    OF    COLUMNS  I  I/ 

The  value  of  the  total  unit-stress  Sc  now  is, 

','( 

*?2\t- 

and  this  is  the  same  as  (10),  except  that  q  has  been  replaced  by 

S    -  -  -- 

A    which  is  not  a  constant  since  it  varies  with  — .    This 

712E  A 

p 

expression  is  a  quadratic  with  reference  to  — ,    and    by   solution 

A 

are  found  the  two  values, 

P  P        7t2Er2 

~A  —  Sgi  anc*  "4  —  — fc™' 

the  first  of  which  corresponds  to  the  formula  for  short  blocks 
and  the  latter  to  Euler's  formula  for  columns  with  round  ends. 

Prob.  103.  Prove  that  /fc'  =  r2  for  a  column  so  deflected  that 
there  is  no  stress  on  the  convex  side,  c'  being  the  distance  from 
that  side  to  the  neutral  axis  of  the  cross-section. 


Il8  ON    TORSION    AND   ON    SHAFTS.  VI. 

I 


CHAPTER  VI. 

ON  TORSION  AND  ON  SHAFTS  FOR  TRANSMITTING  POWER. 
ART.  64.     THE  PHENOMENA  OF  TORSION. 

Torsion  occurs  when  applied  forces  tend  to  cause  a  twisting  of 

a  body  around  an 
axis.  Let  one  end 
of  a  horizontal  shaft 
be  rigidly  fixed  and 
let  the  free  end  have 
a  lever  p  attached  at 

right   angles    to    its 

f 
axis.     A   weight   P 

hung  at  the  end  of 
this  lever  will  twist  the  shaft  so  that  fibers  such  as  ab,  which  were 
originally  horizontal,  assume  a  spiral  form  ad  like  the  strands  of 
a  rope.  Radial  lines  such  as  cb  will  also  have  moved  through  a 
certain  angle  bed. 

Experiments  have  proved,  that  if  P  be  not  so  large  as  to  strain 
the  material  beyond  its  elastic  limit,  the  angles  bed  and  bad 
are  proportional  to  P  and  that  on  the  removal  of  the  stress  the 
lines  cd  and  ad  return  to  their  original  positions  cb  and  ab.  The 
angle  bed  is  evidently  proportional  to  the  length  of  the  shaft, 
while  bad  is  independent  of  the  length.  If  the  elastic  limit  be  ex- 
ceeded this  proportionality  does  not  hold,  and  if  the  twisting  be 
great  enough  the  shaft  will  be  ruptured.  These  laws  are  but  a 
particular  case  of  the  general  axioms  stated  in  Art.  3. 


ART.    65.       THE  FUNDAMENTAL  FORMULA  FOR  TORSION.  I IQ 

The  product  Pp  is  the  moment  of  the  force  P  with  respect  to 
the  axis  of  the  shaft,  p  being  the  perpendicular  distance  from  that 
axis  to  the  line  of  direction  of  P,  and  is  called  the  twisting  mo- 
ment. Whatever  be  the  number  of  forces  acting  at  the  end  of 
the  shaft,  their  resulting  twisting  moment  may  always  be  repre- 
sented by  a  single  product  Pp. 

A  graphical  representation  of  the  phenomena  of  torsion  may 
be  made  as  in  Fig.  I ,  the  angles  of  torsion  being  taken  as  abscis- 
sas and  the  twisting  moments  as  ordinates.  The  curve  is  then  a 
straight  line  from  the  origin  until  the  elastic  limit  of  the  material 
is  reached,  when  a  rapid  change  occurs  and  it  soon  becomes 
nearly  parallel  to  the  axis  of  abscissas.  The  total  angle  of  torr 
sion,  like  the  total  ultimate  elongation,  serves  to  compare  the 
relative  ductility  of  specimens. 

Prob.  104.  A  shaft  2  feet  long  is  twisted  through  an  angle  of  7 
degrees  by  a  force  of  200  pounds  acting  at  a  distance  of  6  inches 
from  the  center.  Through  what  angle  will  a  shaft  4  feet  long  be 
twisted  by  a  force  of  500  pounds  acting  at  a  distance  of  18  inches 
from  the  center? 

ART.  65.     THE  FUNDAMENTAL  FORMULA  FOR  TORSION. 

The  stresses  which  occur  in  torsion  are  those  of  shearing,  each 

cross-section  tending  to  shear  off  from  the  one  adjacent  to  it. 

When  equilibrium  obtains  the  external  twisting  moment  is  exactly 

balanced  by  the  sum  of  the  moments  of  these  internal  stresses,  or, 

Resisting  moment  =  Twisting  moment. 

If  P  be  the  force  acting  at  a  distance  p  from  the  center  about, 
which  the  twisting  takes  place,  the  value  of  the  twisting  moment 
is  Pp.  To  find  the  resisting  moment,  let  c  be  the  distance  from 
the  center  to  the  remotest  part  of  the  cross-section  where  the 
unit-shear  is  ^.  Then  since  the  stresses  vary  as  their  distances 
from  the  center, 

-—  =  unit-stress  at  a  unit's  distance  from  center, 


120  ON   TORSION    AND   ON    SHAFTS.  VI. 

— £—  =  unit-stress  at  a  distance  z  from  center, 
c 

— —  =  total  stress  on  an  elementary  area  a, 

~2 

=  moment  of  this  stress, 

=  internal  resisting  moment. 


c 

This  may  be  written   i  2az*.     But  2az2  is  the  polar  moment  of 

c 

inertia  of  the  cross-section  and  may  be  denoted  byy.    Therefore, 
(11)  ^L=Ppy 

which  is  the  fundamental  formula  for  torsion. 

The  analogy  of  formula  (11)  with  formula  (4)  for  the  flexure 
of  beams  will  be  noted.  Pp,  the  twisting  moment,  is  often  the 
resultant  of  several  forces  and  might  have  been  expressed  by  a 
single  letter  like  the  Min  (4).  By  means  of  (i  i)  a  shaft  subjected 
to  a  given  moment  may  be  investigated,  or  the  proper  size  be 
determined  for  a  shaft  to  resist  given  forces. 

Prob.  105.  A  circular  shaft  is  subjected  to  a  maximum  shearing 
unit-stress  of  2  ooo  pounds  when  twisted  by  a  force  of  90  pounds 
at  a  distance  of  27  inches  from  the  center.  What  unit-stress  will 
be  produced  in  the  same  shaft  by  two  forces  of  40  pounds  one 
acting  at  2 1  and  the  other  at  36  inches  from  the  center  ? 

ART.  66.     POLAR  MOMENTS  OF  INERTIA. 

The  polar  moment  of  inertia  for  simple  figures  is  readily  found 
by  the  help  of  the  calculus,  as  explained  in  works  on  elementary 
mechanics.  It  is  also  a  fudamental  principle  that, 

where  J  is  the  polar  moment  of  inertia,  7t  the  least  and  72  the 
greatest  rectangular  moment  of  inertia  about  two  axes  passing 


ART.    67.  THE    CONSTANTS   OF   TORSION.  121 

through  the  center.  The  following  are  values  of  J  for  some  of 
the  most  common  cases. 

Tid* 
For  a  circle  with  a  diameter  d,  J  = 

For  a  square  whose  side  is  d,  J  = 

6 

For  a  rectangle  with  sides  b  and  d.  J  =  —  _(_ 

12  T  12 

The  value  of  c  in  all  cases  is  the  distance  from  the  center  about 
which  the  twisting  occurs,  usually  the  center  of  figure  of  the 
cross-section,  to  the  remotest  part  of  the  cross-section.  Thus, 

For  a  circle  with  diameter  d,  c  =  ^d, 

For  a  square  whose  side  is  d,  c  =  dy ]/?,. 

For  a  rectangle  with  sides  b  and  d,  c  =  y^y '  b2  +  d2. 

It  is  rare  in  practice  that  formulas  for  torsion  are  needed  for  any 
cross-sections  except  squares  and  circles. 

Prob.  1 06.  Find  the  values  of  J  and  c  for  an  equilateral  tri- 
angle whose  side  is  d. 

ART.  67.     THE  CONSTANTS  OF  TORSION. 

The  constant  ^  computed  from  experiments  on  the  rupture  of 
shafts  by  means  of  formula  (u)  may  be  called  the  modulus  of 
torsion,  in  analogy  with  the  modulus  of  rupture  as  computed 
from  (4).  As  would  be  expected  the  values  thus  found  agree 
closely  with  the  ultimate  shearing  unit-stress  given  in  Art.  7,  viz., 
For  timber,  ^  =  2  ooo  pounds  per  square  inch, 

For  cast  iron,  5,  =  25  ooo  pounds  per  square  inch, 

For  wrought  iron,     ^  =  50  ooo  pounds  per  square  inch, 
For  steel,  S^  =  75  ooo  pounds  per  square  inch. 

By  the  use  of  these  average  values  it  is  hence  easy  to  compute 
from  (i  i)  the  load  P  acting  at  the  distance  /  which  will  cause  the 
rupture  of  a  given  shaft. 


122  ON    TORSION   AND    ON    SHAFTS.  VI. 

The  coefficient  of  elasticity  for  shearing  may  be  computed 
from  experiments  on  torsion  in  the  following  manner.  Let  a  cir- 
cular shaft  whose  length  is  /  and  diameter  d  be  twisted  through 
an  angle  6  by  the  twisting  moment  Pp.  Here  a  point  on  the  cir- 
cumference of  one  end  is  twisted  relative  to  a  corresponding 
point  on  the  other  end  through  the  arc  0  or  through  the  dis- 
tance y2Qd.  From  the  fundamental  definition  of  the  coefficient 
of  elasticity  E  as  given  in  (i), 


s  ~~  Od 
and  inserting  for  ^  its  value  from  (i  i), 

E  -  VPPl 


from  which  E  can  be  computed  when  all  the  quantities  in  the 
second  member  have  been  determined  by  experiment,  provided 
that  the  elastic  limit  of  the  material  be  not  exceeded. 

Prob.  107.  An  iron  shaft  5  feet  long  and  2  inches  in  diameter 
is  twisted  through  an  angle  of  7  degrees  by  a  force  of  5  ooo 
pounds  acting  at  6  inches  from  the  center,  and  on  the  removal  of 
the  force  springs  back  to  its  original  position.  Find  the  value  of 
E  for  shearing. 

Prob.  108.  What  force  P  acting  at  the  end  of  a  lever  24  inches 
long  will  twist  asunder  a  steel  shaft  1.4  inches  in  diameter? 

ART.  68.     SHAFTS  FOR  THE  TRANSMISSION  OF  HORSE  POWER. 

Work  is  the  product  of  a  resistance  by  the  distance  through 
which  it  acts,  and  is  usually  measured  in  foot-pounds.  A  horse- 
power is  33  ooo  foot-pounds  of  work  done  in  one  minute.  It  is 
required  to  determine  the  relation  between  the  horse-power  H 
transmitted  by  a  shaft  and  the  greatest  internal  shearing  unit- 
stress  5^  produced  in  it. 

Let  a  shaft  making  n  revolutions  per  minute  transmit  H  horse- 
power. The  work  may  be  applied  by  a  belt  from  the  motor  to  a 


ART.    69.  ROUND   SHAFTS.  123 

pulley  on  the  shaft,  then,  by  virtue  of  the  elasticity  and  resistance 
of  the  material  of  the  shaft,  it  is  carried  through  other  pulleys  and 
belts  to  the  working  machines.  In  doing  this  the  shaft  is  strained 
and  twisted,  and  evidently  ^  increases  with  H.  Let  P  be 
the  resistance  acting  at  the  circumference  of  the  pulley  and  p 
the  radius  of  the  pulley.  In  making  one  revolution  the  force  P 
acts  through  the  distance  27tp  and  performs  the  work  2itpP,  and 
in  n  revolutions  it  performs  the  work  2xpPn.  Then  if  P  be  in 
pounds  and  p  in  inches  the  imparted  horse-power  is, 
H  =  27lpPn 

33  ooo  X  12 
The  twisting  moment  Pp  in  this  expression  may  be  expressed,  as 


. 
in  formula  (i  i),  by  the  resisting  moment     sJ  .      Hence  the  equa- 

tion becomes, 


198  oooc 

This  is  the  formula  for  the  discussion  of  shafts  for  the  transmis- 
sion of  power,  and  in  it  J  and  c  must  be  taken  in  inches  and  5f 
in  pounds  per  square  inch,  while  n  is  the  number  of  revolutions 
per  minute. 

Prob.  109.  What  horse-power  is  required  to  draw  25  miles  per 
hour  a  train  weighing  400  tons  on  a  track  where  the  coefficient 
of  friction  is  0.006  and  the  grade  30  feet  per  mile  ? 

Prob.  1  10.  A  wooden  shaft  6  inches  square  breaks  when  mak- 
ing 40  revolutions  per  minute.  Find  the  horse-power  then  proba- 
bly transmitted. 

ART.  69.     ROUND  SHAFTS. 

For  round  shafts  of  diameter  d,  the  values  of  J  and  c  are  to  be 
taken  from  Art.  66  and  inserted  in  the  last  equation,  giving, 

St  =  321  ooo  ~        or         d=  68.5-4— 
nffi'  \nS; 

The  first  of  these  may  be  used  for  investigating  the  strength  of 
a  given  shaft  when  transmitting  a  certain  number  of  horse-power 


124  ON    TORSION    AND   ON   SHAFTS.  VI. 

with  a  known  velocity.  The  computed  values  of  Ss,  compared 
with  the  ultimate  values  in  Art.  67,  will  indicate  the  degree  of 
security  of  the  shaft.  Here  d  must  be  taken  in  inches  and  5^  will 
be  in  pounds  per  square  inch. 

The  second  equation  may  be  used  for  determining  the  diameter 
of  a  shaft  to  transmit  a  given  horse-power  with  a  given  number 
of  revolutions  per  minute.  Here  a  safe  allowable  value  must  be 
assumed  for  Ss  in  pounds  per  square  inch,  and  then  d  will  be 
found  in  inches.  This  equation  shows  that  the  diameter  of  a 
shaft  varies  directly  as  the  cube  root  of  the  transmitted  horse- 
power and  inversely  as  the  cube  root  of  its  velocity. 

Prob.  in.  Find  the  factors  of  safety  for  a  wrought  iron  shaft 
2,y2  inches  in  diameter  when  transmitting  25  horse-powers  while 
making  100  revolutions  per  minute,  and  also  when  making  10 
revolutions  per  minute. 

Prob.  112.  Find  the  diameter  of  a  wrought  iron  shaft  to  trans- 
mit 90  horse-powers  with  a  factor  of  safety  of  8  when  making 
250  revolutions  per  minute,  and  also  when  making  100  revolu- 
tions per  minute. 

ART.  70.     SQUARE  SHAFTS. 

For  a  square  shaft  whose  side  is  d  formula  (12)  reduces  to, 
5  =  267500-^,        or       4=64.4  J»l—  • 


These  are  the  same  as  for  round  shafts  except  in  the  numerical 
constants,  and  are  to  be  used  in  the  same  manner,  the  first  to  in- 
vestigate an  existing  shaft  and  the  second  to  find  the  diameter  for 
one  proposed. 

Prob.  113.  Find  the  proper  diameter  of  a  wooden  shaft  for  a 
water  wheel  which  is  to  transmit  8  horse-power  at  20  revolutions 
per  minute. 

Prob.  114.  Find  the  factor  of  safety  of  a  wooden  shaft  12  inches 
square  when  transmitting  16  horse-power  at  40  revolutions  per 
minute. 


ART.  71.  MISCELLANEOUS  EXERCISES.  125 

ART.  71.      MISCELLANEOUS  EXERCISES. 

Exercise  7.  Find  in  the  library  a  description  of  Thurston's 
autographic  testing  machine  for  torsion.  Write  a  condensed  de- 
scription of  it  and  of  the  method  of  its  use.  Give  sketches  of  the 
autographic  recording  apparatus.  Explain  how  the  torsion  dia- 
grams may  be  used  to  study  the  relative  stiffness,  elastic  limit 
and  ductility  of  specimens. 

Exercise  8.  Go  to  a  testing  room  and  inspect  Thurston's  test- 
ing machine  for  torsion.  Ascertain  the  dimensions  and  kind  of 
specimens  tested  thereon.  Explain  with  sketches  the  construction 
of  the  machine  and  the  method  of  its  use.  State  how  the  quality 
of  the  specimens  is  inferred  from  the  torsion  diagrams. 

Exercise  9.  Measure  the  diameter  of  a  shaft,  and  ascertain  its 
velocity  and  the  number  of  transmitted  horse-powers.  State  in 
a  short  report  the  data  and  the  results  of  your  investigations. 

Prob.  115.  Compare  the  strength  of  a  square  shaft  with  that  of 
a  circular  shaft  of  equal  areas. 

Prob.  1  1  6.  Jones  &  Laughlins  give  the  formulas, 

d  =  Jte^tf         and       d  = 

~~ 


the  first  for  ordinary  turned  wrought  iron  shafts,  and  the  second 
for  cold  rolled  wrought  iron  shafts.  What  working  unit-stresses 
do  these  imply  ? 


126  ON    COMBINED    STRESSES.  VII. 


CHAPTER    VII. 

ON  COMBINED   STRESSES. 

ART.  72.     CASES  OF  COMBINED  STRESSES. 

The  three  kinds  of  simple  stress  are  tension,  compression  and 
shear,  or,  in  other  words,  the  numerical  investigation  of  bodies 
under  stress  includes  only  the  unit-stresses  St,  Sc  and  Ss.  Trans- 
verse or  flexural  stress  was  investigated  in  Chapter  III  by  resolv- 
ing the  internal  stresses  into  tension,  compression  and  shear. 
Torsional  stress  is  merely  a  particular  case  of  shear. 

Tension  and  compression  are  similar  in  character  and  differ 
only  in  sign  or  direction.  Hence  their  combination  is  effected  by 
algebraic  addition.  Thus  if  P  be  a  tensile  stress  and  P  a  com- 
pressive  stress  applied  to  the  same  bar  at  the  same  time  the  result- 
ant stress  is  P  —  P  which  may  be  either  tensile  or  compressive. 

Tension  and  shear,  or  compression  and  shear,  are  often  com- 
bined, as  internally  in  the  case  of  beams  and  externally  under 
many  circumstances. 

Tension  and  flexure  are  combined  when  loads  are  placed  upon 
a  bar  under  tension.  This  case  and  that  of  compression  and  flex- 
ure are  of  frequent  occurence,  and  their  investigation  is  of  much 
practical  importance. 

Flexure  and  torsion  are  combined  whenever  shafts  for  the  trans- 
mission of  power  are  loaded  with  pulleys  and  belts,  and,  as  will 
be  seen,  the  effect  of  the  flexure  is  sensibly  to  modify  the  formulas 


ART.    73.  STRESSES    DUE   TO    TEMPERATURE.  12? 

of  the  last  chapter.     Compression  and  flexure  occur  in  the  case 
of  vertical  shafts. 

The  internal  stresses  in  a  body  produced  by  applied  forces  are 
usually  of  a  complex  character.  Even  in  a  case  of  simple  tension 
there  are  shearing  stresses  in  all  directions  except  those  perpen- 
dicular and  parallel  to  the  line  of  tension.  If  P  be  the  tensile 

force  and  A  the  area  of  the  cross-section  of  the  bar  the  tensile 

p 

unit-stress   is  — ,  and  it  may  be  shown,  as  in  Prob.  26,  that  a 
A 

p 

shearing  unit-stress  of  y2 exists  in  a  section  making  an  angle 

A 

of  45  degrees  with  the  axis  of  the  bar. 

Prob.  1 17.  A  pulls  29  pounds  at  one  end  of  a  rope  and  B  pulls 
30  pounds  at  the  other  end.  What  is  the  stress  in  the  rope  ? 

ART.  73.      STRESSES  DUE  TO  TEMPERATURE. 

If  a  bar  be  unstrained  it  expands  when  the  temperature  rises 
and  contracts  when  the  temperature  falls.  But  if  the  bar  be  under 
stress,  so  that  the  change  of  length  cannot  occur,  an  additional 
unit-stress  must  be  produced  which  will  be  equivalent  to  the 
unit-stress  that  would  cause  the  same  change  of  length  in  the 
unstrained  bar.  Thus  if  a  rise  of  temperature  elongates  a  bar  of 
length  unity  the  amount  s  when  free  from  stress,  it  will  cause  the 
unit-stress  S  =  sE  (see  Art.  4)  when  the  bar  is  prevented  from 
expanding  by  external  forces. 

Let  /  be  the  length  of  the  bar,  a  its  coefficient  of  linear  expan- 
sion for  a  change  of  one  degree,  and  X  the  change  of  length  due 
to  the  rise  or  fall  of  t  degrees.  Then, 

I  =  atl, 
and  the  unit-strain  s  is, 


The  unit-stress  produced  by  the  change  in  temperature  hence  is, 
5=  a-tE 


128  ON    COMBINED    STRESSES.  VII. 

which  is  seen  to  be  independent  of  the  length  of  the  bar.     The 
total  stress  on  the  bar  is  then  AS. 

The  following  are  average  values  of  the  coefficients  of  linear 
expansion  for  a  change  in  temperature  of  one  degree  Fahrenheit. 
For  brick  and  stone,  a  =  o.ooo  oo  50, 

For  cast  iron,  a  =  o.ooo  oo  62, 

For  wrought  iron,  a  =  o.ooo  oo  67, 

For  steel.  a  =  o.ooo  oo  65. 

As  an  example  consider  a  wrought  iron  tie  rod  20  feet  in  length 
and  2  inches  in  diameter  which  is  screwed  up  to  a  tension  of 
9  ooo  pounds  in  order  to  tie  together  two  walls  of  a  building. 
Let  it  be  required  to  find  the  stress  in  the  rod  when  the  tempera- 
ture falls  10°  F.  Here, 

5  =  o.ooo  oo  67  X  10  X  25  ooo  ooo  =  i  675  pounds. 
The  total  tension  in  the  rod  now  is, 

9  ooo  -f-  3.14  X  i  675  =  14  ooo  pounds. 
Should  the  temperature  rise  10°  the  tension  in  the  rod  would  be, 

9  ooo  —  3.14  X  i  675  =  4  ooo  pounds. 

In  all  cases  the  stresses  caused  by  temperature  are  added  or 
substracted  to  the  tensile  or  compressive  stresses  already  existing. 

Prob.  1 1 8.  A  cast  iron  bar  is  confined  between  two  immovable 
walls.  What  unit-stress  will  be  produced  by  a  rise  of  40°  in 
temperature  ? 

ART.  74.     COMBINED  TENSION  AND  FLEXURE. 
Consider  a  beam  in  which  the  flexure  produces  a  unit-stress  £ 
at  the  fiber  on  the  tensile  side  most  remote  from  the  neutral  axis. 
Let  a  tensile  stress  P  be  then  applied  to  the  ends  of  the  bar  uni- 
formly distributed  over  the  cross-section  A.     The  tensile  unit- 

p 
stress  at  the  neutral  surface  is  then  —  and  all  the    longitudinal 

A 

stresses  due  to  the  flexure  are  increased  by  this  amount.     The 

p 

maximum  tensile  unit-stress  is  then    +  S  in  which  j>  is  to  be 

A 

found  from  formula  (4). 


ART.    75.  COMBINED    COMPRESSION    AND    FLEXURE.  I2Q 

In  designing  a  beam  under  combined  tension  and  flexure  the 

p 
dimensions  must  be  so  chosen  that  —  +  -S"  shall  not  exceed  the 

A 

proper  allowable  working  unit-stress.  For  instance,  let  it  be  re- 
quired to  find  the  size  of  a  square  wooden  beam  of  12  feet  span 
to  hold  a  load  of  300  pounds  at  the  middle  while  under  a  longi- 
tudinal stress  of  2  ooo  pounds,  so  that  the  maximum  tensile  unit- 
stress  may  be  about  I  ooo  pounds  per  square  inch.  Let  d  be  the 
side  of  the  square.  From  formula  (4), 

^  __  6M  _  6  X  150  X  72 

~W  ~          ~d~ 

Then  from  the  conditions  of  the  problem, 
2000+6480^; 

d?  d* 

from  which  results  the  cubic  equation, 

J3  —  2d  =  64.8, 
whose  solution  gives  for  d  the  value  4.25  inches. 

In  investigating  a  beam  under  combined  tension  and  flexure 

p 

the  maximum  value  of  —  +  6"  is  to  be  computed,  and  the  factor 
A 

of  safety  found  by  comparing  it  with  the  ultimate  tensile  strength 
of  the  material. 

Prob.  119.  A  heavy  12-inch  I  beam  carries  a  uniform  load  of 
200  pounds  per  linear  foot,  besides  its  own  weight,  and  is  sub- 
jected to  a  longitudinal  tension  of  80  ooo  pounds.  Find  the 
factor  of  safety  of  the  beam. 

Prob.  1 20.  What  I  beam  is  required  to  carry  a  uniform  load 
of  200  pounds  per  linear  foot  when  subjected  to  a  tension  of  50  ooo 
pounds,  the  maximum  tensile  stress  to  be  9  ooo  pounds  per 
square  inch? 

ART.  75.     COMBINED  COMPRESSION  AND  FLEXURE. 

Consider  a  beam  in  which  the  flexure  produces  a  unit-stress  5" 
in  the  fiber  on  the  compressive  side  most  remote  from  the  neutral 


I3O  ON    COMBINED    STRESSES.  VII. 

axis.     Let  a  compressive  stress  P  be  applied  in  the  direction  of 

its  length  uniformly  over  the  cross-section  A.     Then  at  the  neu- 

p 
tral  surface  the  unit-stress  is  — -  and  at  the  remotest  fiber  it  is 

A 
p 

—  +  S.  The  discussion  of  this  case  is  hence  exactly  similar  to 
A 

that  of  the  last  article.  If  the  beam  is  short  the  total  working 
unit-stress  is  to  be  taken  as  for  a  short  prism ;  if  long  it  should 
be  derived  from  Gordon's  formula  for  columns. 

The  method  of  investigation  explained  in  this  and  the  preced- 
ing article  is  the  one  ordinarily  used  in  practice  on  account  of  the 
complexity  of  the  formulas  which  result  from  the  strict  mathe- 
matical determination  of  the  moments  of  the  applied  forces.  Al- 
though not  exact  the  method  closely  approximates  to  the  truth, 
giving  values  of  the  stresses  a  little  too  large  for  the  case  of  ten- 
sion and  a  little  too  small  for  the  case  of  compression. 

An  inclined  beam  is  an  instance  of  combined  flexure  and  com- 
pression. In  the  case  shown  in 
Fig.  47  the  reactions  are  vertical 
and  their  values  for  any  given 
loads  are  found  by  the  principles 
of  Art.  14.  Let  <p  be  the  inclina- 
tion of  the  beam  to  the  vertical, 
F1&47'  and  for  illustration  let  the  load  be 

uniform.  Then  Rt  =  R2  —  %wl,  \iw  is  the  load  per  linear  unit. 
At  any  section  x  the  unit-stress  5  due  to  the  flexure  is  added  to 
the  compressive  unit-stress  Se  due  to  the  components  of  R^  and 
wx  which  are  parallel  to  the  beam.  Thus  for  a  rectangular  beam 
whose  breadth  is  b  and  depth  d  the  formula  (4)  gives, 

<-. 6M ytvlx  cos  (p  —  yvx*  cos  <p 

~~bd*   ~  bd* 

while  from  (i)  the  direct  compression  is, 

<-.  wx  sin  <p  —  RI  sin  <p 

~~ 


ART.    75.  COMBINED    COMPRESSION    AND    FLEXURE. 


The  total  compressive  unit-stress  at  the  fiber  on  the  upper  side 
now  is, 

s,  =  s  +  se  =  ^1?  (&  --*•)  +  ZJJLS  (2*  -  1). 

It  is  easy  to  show  that  this  is  a  maximum  when, 
_  /     ,    d  tan  <f> 
~  2  "•    ~6~ 
and  therefore  the  maximum  unit-stress  is, 

-     _  3w/2  cos  <p         w  sin  <p  tan  <p 
~~ 


This  equation  may  be  used  to  determine  the  factor  of  safety  of  a 
given  beam  or  to  design  one  proposed. 

A  rafter  of  a  roof  is  a  case  where  one  of  the  reactions  of  the  in- 
clined beam  is  horizon- 
tal as  shown  in  Fig.  48. 
If  /  be  the  length,  w  the 
load  per  linear  unit  and 
<p  the  inclination,  H  is 
to  be  found  by  Art  14  ; 
thus  taking  the  cen- 
ter of  moments  at  the 
lower  end, 


H  .  I  sin  <p  = 


/C°S 


whence 


TT 

H  =  —  cot  <p, 


For  any  section  x,  the  flexural  unit-stress  now  is, 

<- 6(Hx  sin  <p  —  y^wx*  cos  <p) 

bd*  ~ 
and  the  uniform  compressive  unit-stress  is, 

<-.  H  cos  (p  -f  wx  sin  <f> 

~~bd~ 
The  total  compressive  unit-stress  on  the  upper  fiber  hence  is, 


c  = 


bd* 


i  wx  sn  y 

bd 


132  ON    COMBINED   STRESSES.  VII. 

This  is  a  maximum  when  x  has  the  same  value  as  for  the  last 
case,  and, 

<-.      _  3o>/2  cos  <p        wl  cosec  <p       w  sin  <p  tan  <p 
~          ~2bd~ 


is  the  greatest  compressive  unit-stress. 

In  any  inclined  rafter  let  P  denote  all  the  load  above  a  section 
distant  x  from  the  upper  end.  Then  the  greatest  unit-stress  for 
that  section  is, 

c    _  Me        P  sin  tf>        H  cos  <p 
^~T^     ~A~        ~A~ 
from  which  Sw  may  be  found  for  any  given  case. 

Prob.  121.  A  wooden  beam  10  inches  wide  and  8  feet  long 
carries  a  uniform  load  of  500  pounds  per  linear  foot  and  is  sub- 
jected to  a  longitudinal  compression  of  40  ooo  pounds.  Find  the 
depth  of  the  beam  so  that  the  maximum  working  unit-stress  may 
be  about  800  pounds  per  square  inch. 

Prob.  122.  A  roof  with  two  equal  rafters  is  40  feet  in  span  and 
15  feet  in  height.  The  wooden  rafters  are  4  inches  wide,  6 
inches  deep  and  carry  450  pounds  per  linear  foot.  Find  the  fac- 
tor of  safety  of  the  rafters. 

Prob.  123.  A  roof  with  two  equal  rafters  is  40  feet  in  span  and 
1  5  feet  in  height.  The  wooden  rafters  are  4  inches  wide  and  each 
carries  a  load  of  450  pounds  at  the  center.  Find  the  depth  of  the 
rafter  so  that  Sm  may  be  700  pounds  per  square  inch. 

ART.  76.     SHEAR  COMBINED  WITH  TENSION  OR  COMPRESSION. 

Let  a  bar  whose  cross-section  is  A  be  subjected  to  the  longi- 
tudinal tension  or  compression  P  and  at  the  same  time  to  a  shear 
V  at  right  angles  to  its  length.  The  longitudinal  unit-stress  is 

P  V 

_  which  may  be  denoted  by  /,  and  the  shearing  unit-stress  is  — 

A  A 

which  may  be  denoted  by  v.  It  is  required  to  find  the  maximum 
unit-stresses  produced  by  the  combination  of/  and  v.  In  the  fol- 
lowing demonstration  P  will  be  regarded  as  a  tensile  force, 


ART    76.  SHEAR    WITH    TENSION    OR    COMPRESSION.  133 

although  the  reasoning  and  conclusions  apply  equally  well  when 
it  is  compressive. 

Consider  an  elementary  cubic  particle  with  edges  one  unit  in 
length  acted  upon  by  the  horizontal  tensile  force  p  and  p,  and  by 
the  vertical  shear  v  and  v,  as  shown  in 

If 

Fig.  49.  These  forces  are  not  in  equilib- 
rium unless  a  horizontal  couple  be  ap- 
plied as  in  the  figure,  each  of  whose  forces 
is  equal  to  v.  Therefore  at  every  point 
of  a  body  under  vertical  shear  there  ex- 
ists a  horizontal  shear,  and  the  horizon- 
tal shearing  unit-stress  is  equal  to  the  vertical  shearing  unit-stress. 

Let  a  parallelopipedal  element  have  the  length  dm,  the  height 
dn  and  a  width  of  unity. 
The  tensile  force  pdn  tends 
to  pull  it  apart  longitudinally. 
The  vertical  shear  vdn  tends 
to  cause  rotation  and  this  is 
resisted,  as  shown  above,  by 

the  horizontal  shear  vdm.  These  forces  may  be  resolved  into 
rectangular  components  parallel  and  perpendicular  to  the  diagonal 
dz,  as  shown  in  Fig.  50.  The  components  parallel  to  the  diago- 
nal form  a  shearing  force  sdz,  and  those  perpendicular  to  it  a  ten- 
sile force  tdz,  s  being  the  shearing  and  /  the  tensile  unit-stresses. 
Let  (p  be  the  angle  between  dz  and  dm.  The  problem  is  first  to 
state  expressions  for  sdz  and  tdz  in  terms  of  <p,  and  then  to  de- 
termine the  value  of  <p,  or  the  ratio  of  dm  to  dn,  which  gives  the 
maximum  values  of  s  and  /. 

By  simple  resolution  of  forces, 

sdz  —  pdn  cos  <p  -\-  vdm  cos  <p  —  vdn  sin  <p, 
tdz  =  pdn  sin  <p  -f-  vdm  sin  <p  -}-  vdn  cos  <p. 

Divide  each  of  these  by  dz,  for  —  put  its  value  sin  <f>  and  for  — 

dz  dz 


134  ON    COMBINED   STRESSES.  VII. 

its  value  cos  <p.     Then  the  equations  take  the  form, 

s  =  p  sin  (p  cos  <p' -\-  z/(cos2  <p  —  sin2  <p), 
t  =  p  sin2  <p  +  2v  sin  <p  cos  <p. 
These  may  be  written, 

s  =  y^t  sin  2<p  +  s  cos  2<f>, 

t  =  y2t  (i  —  cos  2<f>)  -f-  j  sin  2^>. 

By  placing  the  first  derivative  of  each  of  these  equal  to  zero  it 
is  found  that, 

s  is  a  maximum  when   tan  2<p  =  —, 

2V 

,  2V 

t  is  a  maximum  when   tan  2<p  = 

A 

Expressing  sin  2(f>  and  cos  2<f>  in  terms  of  tan  2(p  and  inserting 
them  in  the  above  the  following  values  result, 


max  s  =  ± 

<">  / 

max  t  =   — 

4' 

These  formulas  apply  to  the  discussion  of  the  internal  stresses 
in  beams,  as  well  as  to  combined  longitudinal  stress  and  vertical 
shear  directly  applied  by  external  forces.  If/  is  tension  t  is  ten- 
sion ;  if  p  is  compression  /  is  also  compression. 

Prob.  124.  A  rivet  ^-inch  in  diameter  is  subjected  to  a  tension 
of  2  ooo  pounds  and  at  the  same  time  to  a  cross  shear  of  3  ooo 
pounds.  Find  the  maximum  tensile  and  shearing  unit-stresses 
and  the  directions  they  make  with  the  axis  of  the  rivet. 

ART.  77..    COMBINED  FLEXURE  AND  TORSION. 

This  case  occurs  when  a  shaft  for  the  transmission  of  power  is 
loaded  with  weights.  Let  vS  be  the  greatest  flexural  unit-stress 
computed  from  (4)  and  5^  the  torsional  shearing  unit-stress  com- 
puted from  (12)  or  by  the  special  equations  of  Arts.  68  and  69. 


ART.    77.  COMBINED   FLEXURE  AND   TORSION.  135 

Then,  according  to  the  last  article  the  resultant  maximum  unit- 
stresses  are, 

6"  /  ^2 

max.  ten.  or  comp.  /  =  -  _j- .  I  .$/  -f-  - 
max  shear  s 


=  ±      I  $•  -f  __ 

>f    '          4 


For  wrought  iron  or  steel  it  is  usually  necessary  to  regard  only 
the  first  of  these  unit-stresses,  but  for  timber  the  second  should 
also  be  kept  in  view. 

For  example,  let  it  be  required  to  find  the  factor  of  safety  of  a 
wrought  iron  shaft  3  inches  in  diameter  and  1 2  feet  between  bear- 
ings, which  transmits  40  horse-power  while  making  120  revolu- 
tions per  minute,  and  upon  which  a  load  of  800  pounds  is  brought 
by  a  belt  and  pulley  at  the  middle.  Taking  the  shaft  as  fixed 
over  the  bearings  the  flexural  unit-stress  is, 

API 

6  =  ^-—  =  5  400  pounds  per  square  inch. 
From  Art.  68  the  torsional  unit-stress  is, 

T  T 

Ss  =321  ooo  — —  —  4  ooo  pounds  per  square  inch. 

The  maximum  tensile  and  compressive  unit-stress  now  is, 

t=  2  700  -f-  y '4  ooo2  -f  2  7OO2  =  7  600  pounds  per  square  in. 

and  the  factor  of  safety  is  hence  over  7. 

As  a  second  example,  let  it  be  required  to  find  the  size  of 
a  square  wooden  shaft  for  a  water  wheel  weighing  3  ooo  pounds 
which  transmits  8  horse-power  while  making  20  revolutions  per 
minute.  The  length  of  the  shaft  is  16  feet  and  one-third  of  the 
weight  is  concentrated  at  the  center  and  the  remainder  is  equally 
divided  between  two  points  each  6  feet  from  the  center.  Here 
the  greatest  flexural  unit-stress  is, 

5  =  6(r  SOP  X  96  —  i  ooo  X  72)  _  49  200 
d~  '  ~      d*     ' 


136  ON    COMBINED   STRESSES.  VII. 

and  from  Art.  69  the  torsional  unit-stress  is, 

_  267  500  X  8    __  107  ooo 
*  ~  ~~~~ 


From  the  formula  of  the  last  article  the  combined  tensile  or  com- 
pressive  stress  is, 

f=  I3440Q. 

fc 

Now  if  the  working  value  of  t  be  taken  at  600  pounds  per  square 
inch  the  value  of  d  will  be  about  6  inches.  From  formula  (  1  3)  also 
_  109  800 

fc 

and  if  the  working  value  of  s  be  taken  at  1  50,  the  value  of  d  is 
found  to  be  about  9  inches.  The  latter  value  should  hence  be 
chosen  for  the  size  of  the  shaft. 

Prob.  125.  Prove  that  the  formula  for  finding  the  diameter  of  a 
round  iron  shaft  is, 


l6M       —    \M*      4°2 

"    t  \  7T2   H 


Kt  t\7f  n 

where  M  is  the  maximum  bending  moment  of  the  transverse 
forces  in  pound-inches,  H  the  number  of  transmitted  horse-power, 
n  the  number  of  revolutions  per  minute,  and  t  the  safe  allowable 
tensile  or  compressive  working  strength  of  the  material. 

ART.  78.     COMBINED  COMPRESSION  AND  TORSION, 

In  the  case  of  a  vertical  shaft  the  torsional  unit-stress  St  com- 
bines with  the  direct  compressive  stress  due  to  the  weights  upon 
the  shaft,  and  produces  a  resultant  compression  t  and  shear  s. 
From  formulas  (13)  the  combined  stresses  are, 


*=+> 

4 

The  use  of  these  is  the  same  as  those  of  the  last  article,  5f  being 


ART.    78.  COMBINED    COMPRESSION    AND   TORSION.  137 

found  from  the  formulas  of  chapter  VI,  while  Sc  is  computed  from 
formula  (i)  if  the  length  of  the  shaft  be  less  than  ten  times  its 
diameter  and  from  (10)  for  greater  lengths. 

Prob.  1 26.  A  vertical  shaft,  weighing  with  its  loads  6  ooo 
pounds,  is  subjected  to  a  twisting  moment  by  a  force  of  300 
pounds  acting  at  a  distance  of  4  feet  from  its  center.  If  the  shaft 
is  wrought  iron,  4  feet  long  and  2  inches  in  diameter,  find  its 
factor  of  safety. 

Prob.  1 27.  Find  the  diameter  of  a  short  vertical  steel  shaft  to 
carry  loads  amounting  to  6  ooo  pounds  when  twisted  by  a  force 
of  300  pounds  acting  at  a  distance  of  4  feet  from  the  center,  taking 
the  unit-stress  against  tension  as  10  ooo  and  against  shearing  as 
7  ooo  pounds  per  square  inch. 


138  APPENDIX    AND   TABLES.  VIII. 


CHAPTER    VIII. 

APPENDIX  AND  TABLES. 

ART.  79.     HORIZONTAL  SHEAR  IN  BEAMS. 

The  common  theory  of  flexure  as  presented  in  Chapters  III  and 
IV  considers  that  the  internal  stresses  at  any  section  are  resolved 
into  their  horizontal  and  vertical  components,  the  former  produc- 
ing longitudinal  tension  and  compression  and  the  latter  a  trans- 
verse shear,  and  that  these  act  independently  of  each  other. 
Formula  (3)  supposes  further  that  the  vertical  shear  is  uniformly 
distributed  over  the  cross-section  of  the  beam.  A  closer  analysis 
will  show  that  a  horizontal  shear  exists  also  and  that  this, 
together  with  the  vertical  shear,  varies  in  intensity  from  the  neu- 
tral surface  to  the  upper  and  lower  sides  of  the  beam.  It  is  well- 
known  that  a  pile  of  boards  which  acts  like  a  beam  deflects  more 
than  a  solid  timber  of  the  same  depth,  and  this  is  largely  due  to 
the  lack  of  horizontal  resistance  between  the  layers.  The  com- 
mon theory  of  flexure  in  neglecting  the  horizontal  shear  gener- 
ally errs  on  the  side  of  safety.  In  a  few  experiments  however 
beams  have  been  known  to  crack  along  the  neutral  surface  and  it 
is  hence  desirable  to  investigate  the  effect  of  horizontal  shear  in 
tending  to  cause  rupture  of  that  kind.  That  a  horizontal  shear 
exists  simultaneously  with  the  vertical  shear  is  evident  from  the 
considerations  in  Art.  76. 

Let  Fig.  5 1  represent  a  portion  of  a  bent  beam  of  uniform  sec- 
tion. Let  a  rectangular  notch  nmpq  be  imagined  to  be  cut  into 


ART.  79. 


HORIZONTAL   SHEAR    IN    BEAMS. 


139 


it,  and  let  forces  be  applied  to  it  to  preserve  the  equilibrium. 
Let   H   be    the 
sum   of  all   the 
horizontal  com-         ' 
ponents  of  these 
forces  acting  on 
mn  and  H'  the 
sum  of  those  act- 
ing on  qp.    Now 

H'  is  greater  or  less  than  //,  hence  the  difference  H'  —  H  must 
act  along  mq  as  a  horizontal  shear.  Let  the  distance  mq  be  dx, 
the  thickness  mm1  be  b,  and  the  area  mqmm'  be  at  a  distance  c' 
above  the  neutral  surface.  Let  c  be  the  distance  from  that 
neutral  surface  to  the  remotest  fiber  where  the  unit-stress  is  6". 
Let  a  be  the  cross-section  of  any  fiber.  Let  M  be  the  bending 
moment  at  the  section  mn  and  M'  that  at  the  section  qp.  Now 
from  the  fundamental  laws  of  flexure, 

—  =    unit-stress  at  a  unit's  distance  from  neutral  surface. 


=   unit-stress  at  distance  y  from  neutral  surface, 


-y 
a  y  =   total  stress  on  fiber  a  at  distance  y, 


=  sum  of  horizontal  stresses  between  m  and  n. 


The  value  of  H  hence 


is,  since   —  —  —  , 
c  I 


and  likewise, 

The  horizontal  shear  therefore  is, 

if<-ff=*^. 


I4O  APPENDIX    AND   TABLES.  VIII. 

Now  since  the  distance  mq  is  dx,  the  value  of  M'  —  M  is  dM. 
Also  if  Sh  be  the  horizontal  shearing  unit-stress  upon  the  area 
bdx  the  value  of  H'  —  H  is  Shbdx.  Hence, 


Again  from  Art.  41  it  is  plain  that  -  is   the  vertical  shear  V 
at  the  section  under  consideration.     Therefore, 


is  the  formula  for  the  horizontal  unit-shear  at  any  point  of  the 
beam. 

This  expression  shows  that  the  horizontal  unit-shear  is  great- 
est at  the  supports,  and  zero  at  the  dangerous  section  where  Fis 
zero.  The  summation  expression  is  the  statical  moment  of  the 
area  mm'nn'  with  reference  to  the  neutral  axis  ;  it  is  zero  when 
y  =  c,  and  a  maximum  when  y  =  o.  Hence  the  longitudinal 
unit-shear  Sh  is  zero  at  the  upper  and  lower  sides  of  the  beam 
and  is  a  maximum  at  the  neutral  surface.  The  formula  for  the 
maximum  horizontal  unit-shear  therefore  is, 

max  V    c 
St  =  —f—Z,<y. 

Here  /  is  the  moment  of  inertia  of  the  whole  cross-section  with 
reference  to  the  neutral  axis  (Art.  22),  b  is  the  width  of  the  beam 
along  the  neutral  surface,  and  3?0  ay  is  the  statical  moment  of 
the  area  of  the  part  of  the  cross-section  on  one  side  of  the 
neutral  axis. 

For  a  rectangular  beam  of  breadth  b  and  depth  d,  the  value  of 

.    bd*        .  „«  bd  d        bd*     „. 

f  is  —  ,  and  2,  ay  =  —  -  —  —  .     Then, 

12  24 

5   -  ^ 
Sh~  -M' 

By  inserting  in  this  the  values  of  V  for  particular  sections  the 
corresponding  values  of  Sh  are  found. 


ART.    8O.        MAXIMUM    INTERNAL   STRESSES    IN    BEAMS.  14! 

Prob.  1  28.  Prove  that  for  a  cylindrical  beam  of  diameter  d  the 

horizontal  unit-shear  along  the  neutral  surface  is  J  - 

3^2' 

Prob.  1  29.  In  the  Journal  of  the  Franklin  Institute  for  Febru- 
ary, 1883,  is  detailed  an  experiment  on  a  spruce  joist  3^X12 
inches  and  14  feet  long,  which  broke  by  tension  at  the  middle 
and  afterwards  by  shearing  along  the  neutral  axis  at  the  end 
when  loaded  at  the  middle  with  12  545  pounds.  Find  the  tensile 
and  shearing  unit-stresses. 

ART.  80.     MAXIMUM  INTERNAL  STRESSES  IN  BEAMS. 

From  the  last  article  it  is  evident  that  at  every  point  of  a  beam 
there  exists  a  horizontal  unit-shear  of  the  intensity  Sh  and  also  a 
vertical  unit-shear  of  the  same  intensity,  whose  value  is  given  by 
(14).  At  every  point  there  also  exists  a  longitudinal  tension  or 
compression  which  may  be  computed  from  (4)  with  the  aid  of  the 
principle  that  these  stresses  vary  directly  as  their  distances  from 
the  neutral  axis.  Let  v  denote  the  unit-shear  thus  determined 
and  p  the  tensile  or  compressive  unit-stress.  Then  from  Art.  76 
the  maximum  unit-shear  at  that  point  is, 


and  it  makes  an  angle  (p  with  the  neutral  surface  such  that, 

•h 

tan  2<p  =  —  • 

2V 

Also  the  tensile  or  compressive  unit-stress  at  that  point  is, 


and  it  makes  an  angle  6  with  the  neutral  surface  such  that, 


From  these  formulas  the  lines  of  direction  of  the   maximum 
stresses  may  be  traced  throughout  the  beam. 


142  APPENDIX   AND   TABLES.  VIII. 

For  the  maximum  shear  v  is  greatest  and  p  is  zero  at  the  neu- 
tral surface,  while  v  is  zero  and  p  is  greatest  at  the  upper  and 
lower  surfaces.  Hence  for  the  neutral  surface  <p  is  o,  it  increases 
with/,  and  becomes  45°  at  the  upper  and  lower  surfaces. 

For  the  maximum  tension  /  is  greatest  and  equal  to  p  on  the 
convex  side  where  v  =  o  and  6  =  0.  As  the  neutral  surface  is 
approached  v  increases,  p  decreases,  and  6  increases.  At  the  neu- 
tral surface  v  is  greatest,  p  is  zero,  and  6  = —  45°.  Here  the 
maximum  tension  and  compression  are  each  equal  to  v. 

For  the  maximum  compression  in  like  manner  6  is  o°  at  the 
concave  surface  and  45°  at  the  neutral  surface.  The  lines  of  max- 
imum tension  if  produced  beyond  the  neutral  surface  would  evi- 
dently cut  those  of  maximum  compression  at  right  angles  and  be 
vertical  at  the  concave  surface. 

The  following  figure  is  an  attempt  to  represent  the  lines  of 

maximum  stress  in 
a  beam.  The  full 
lines  above  the  neu- 
tral surface  are  those 
of  maximum  com- 
pression, while  those 
below  are  maximum 
tension.  The  broken 
lines  are  those  of 
maximum  shear.  On 

any  line  the  intensity  of  stress  varies  with  the  inclination,  being 
greatest  where  the  line  is  horizontal  and  least  where  its  inclina- 
tion is  45°.  The  lines  of  maximum  shear  cut  those  of  maximum 
tension  and  compression  at  angles  of  45°.  The  lines  of  maxi- 
mum tension  above  the  neutral  surface  and  those  of  maximum 
compression  below  it  are  not  shown. 

It  appears  from  the  investigation  that  the  common  theory  of 
flexure  gives  the  horizontal  unit-stress  correctly  at  the  dangerous 


ART.    8 1.  THE   FATIGUE  OF  MATERIALS.  143 

section  of  a  simple  beam  where  the  vertical  shear  is  zero.  At  other 
sections  the  stress  5  as  computed  from  (4)  is  correct  for  the  re- 
motest fiber,  but  for  other  fibers  the  unit-stress  t  is  greater.  It  is 
hence  seen  that  the  main  practical  value  of  the  theory  of  internal 
stress  is  in  showing  that  the  intensity  of  the  shear  varies  through- 
out the  cross-section  of  the  beam.  For  a  restrained  beam,  where 
the  vertical  shear  suddenly  changes  sign  at  the  dangerous 
section,  the  common  theory  gives  the  horizontal  stress  5"  correctly 
for  the  remotest  fiber  only,  and  it  might  be  possible  for  the  maxi- 
mum stress  t  to  be  greater  than  5"  for  a  fiber  nearer  to  the  neutral 
surface. 

Prob.  130.  A  joist  fixed  at  both  ends  is  3  X  12  inches  and  12 
feet  long,  and  is  strained  by  a  load  at  the  middle,  so  that  the  value 
of  5"  as  computed  from  (4)  is  4  ooo  pounds  per  square  inch.  Find 
the  value  of  /  for  points  over  the  support  distant  3,4  and  5  inches 
from  the  neutral  surface. 

ART.  Si.     THE  FATIGUE  OF  MATERIALS. 

The  ultimate  strength  Su  is  usually  understood  to  be  that  unit- 
stress  which  causes  rupture  at  one  application.  Experience  and 
experiments,  however,  teach  that  if  a  unit-stress  somewhat  less 
than  Su  be  applied  a  sufficient  number  of  times  to  a  bar  rupture 
will  be  caused.  The  experiments  of  Wohler  have  been  of  the 
greatest  value  in  establishing  the  laws  which  govern  the  rupture 
of  metals  under  repeated  applications  of  stress.  For  instance,  he 
found  that  the  rupture  of  a  bar  of  wrought  iron  by  tension  was 
caused  in  the  following  different  ways. 

By  800  applications  of  5  2  800  pounds  per  square  inch. 

By         107  ooo  applications  of  48  400  pounds  per  square  inch. 

By         450  ooo  applications  of  39  ooo  pounds  per  square  inch. 

By  10  140  ooo  applications  of  35  ooo  pounds  per  square  inch. 
The  range  of  stress  in  each  of  these  applications  was  from  o  to 
the  designated  number  of  pounds  per  square  inch.  Here  it  is  seen 


144  APPENDIX    AND   TABLES.  VIII. 

that  the  breaking  stress  decreases  as  the  number  of  applications 
increase.  In  other  experiments  where  the  initial  stress  was  not  o, 
but  a  permanent  value  S,  the  same  law  was  seen  to  hold  good. 
It  was  further  observed  that  a  bar  could  be  strained  from  o  up  to 
its  elastic  limit  an  enormous  number  of  times  without  rupture. 
From  a  discussion  of  his  numerous  experiments  Wohler  stated 
the  following  laws. 

1.  By  repeated  applications  of  stress  rupture  may  be  caused 
by  a  unit-stress  less  in  value  than  the  ultimate  strength 
of  the  material. 

2.  The  greater  the  range  of  stress  the  less  is  the  unit-stress 
required  to  produce  rupture  after  an  enormous  number  of 
applications. 

3.  When  the  stress  ranges  from  o  up  to  a  value  about  equal 
to  the  elastic  limit  the  number  of  applications  required  to 
rupture  it  is  enormous,  or  greater  than  could  be  applied 
in  practice. 

4.  A  range  of  stress  from  tension  into  compression,  or  vice 
versa,  produces  rupture  sooner  than  the  same  range  in 
stress  of  one  kind  only. 

5.  When  the  range  of  stress  in  tension  is  equal  to  that  in 
compression  the  stress  which  will  produce  rupture  after 
an  enormous  number  of  applications  is  a  little  greater 
than  one-half  the  elastic  limit. 

The  term  '  enormous  number '  used  in  stating  these  laws  means 
about  40  millions,  that  being  roughly  the  number  used  by  Woh- 
ler to  cause  rupture  under  the  conditions  stated.  For  all  practi- 
cal cases  of  repeated  stress,  except  in  fast  moving  machinery,  this 
great  number  would  seldom  be  exceeded  during  the  natural  life 
of  the  piece. 

In  Art.  8  it  was  recognized  that  the  working  strength  should 
be  less  for  pieces  subject  to  varying  stresses  than  for  those  carry- 
ing steady  loads  only.  For  many  years  indeed  it  has  been  the 


ART.    82.     WORKING  STRENGTH  FOR  REPEATED  STRESSES.  145 

practice  of  designers  to  grade  the  working  strength  according  to 
the  range  of  stresses  to  which  it  might  be  liable  to  be  subjected, 
Wohler's  laws  and  experiments  afford  however  a  means  of 
grading  these  values  in  a  more  satisfactory  manner  than  mere 
judgment  can  do,  and  a  formula  for  that  purpose  will  be  deduced 
in  the  next  article.  After  the  working  strength  Sw  is  determined 
the  cross-section  of  the  piece  is  found  in  the  usual  way,  if  in 
tension  by  formula  (i),  and  if  in  compression  by  formula  (i)  or 
(10)  as  the  case  may  require. 

Prob.  131.  How  many  years  will  probably  be  required  for  a  tie 
bar  in  a  bridge  truss  to  receive  40  million  repetitions  of  stress  ? 

ART.  82.     WORKING  STRENGTH  FOR  REPEATED  STRESSES. 

Consider  a  bar  in  which  the  unit-stress  varies  from  Sf  to  S, 
the  latter  being  the  greater  numerically.  Both  S'  and  S  may  be 
tension  or  both  may  be  compression,  or  one  may  be  tension  and 
the  other  compression.  In  the  last  case  the  sign  of  Sf  is  to  be 
taken  as  minus.  Consider  the  stress  to  be  repeated  an  enormous 
number  of  times  and  rupture  to  then  occur.  By  Wohler's  second 
law  .S  is  some  function  of  the  range  of  stress,  or, 

s  =  <;>(s-s>). 

This  may  be  expressed  in  another  way,  thus, 


or,  in  words,  the  rupturing  stress  S  after  an  enormous  number  of 
repetitions  is  a  function  of  the  ratio  of  the  limiting  stresses. 

Let  u  be  the  ultimate  strength  of  the  material,  tensile  if  >S  is 
tension  and  compressive  if  5  is  compression.  Let  e  be  the  unit- 
stress  at  the  elastic  limit,  and  f  the  unit-stress  which  produces 
rupture  after  an  enormous  number  of  repetitions  when  the  range 
of  stress  in  tension  is  equal  to  that  in  compression.  It  is  required 

to  find  the  value  of  S  in  terms  of  u,  e,  f  and  the  ratio  —  .      For 

O 


146  APPENDIX    AND   TABLES.  VIII. 

this  purpose  assume  the  function, 


in  which  m,  n  and  p  are  quantities  to  be  determined.  Now  if 
S'  =  S,  there  is  no  range  of  stress,  and  the  case  corresponds  to 
that  of  a  steady  load  for  which  6"  =  ti.  Again,  let  S'  =  o,  then 
by  Wohler's  third  law  S  =  e.  Lastly,  let  S'  =  —  S,  then  by 
Wohler's  fifth  law  the  value  of  6"  is  f.  For  these  three  condi- 
tions the  assumed  function  becomes, 

For  S'  =  S,  u  =  me  -\-n-\-p, 

For  S'  =  o,  e  =  me, 

For  S'  =  —  S,      f=  me  —  n  +  p. 
From  these  three  equations  are  found  the  values, 

u  —  /  u  +  f  —  2.e 

m  =  i,  n  =  -  -,  p  =  —^  --  . 

2  2 

and  the  expression  for  S  hence  is, 


This  formula  is  not  to  be  regarded  as  the  true  law  of  rupturing 
strength  under  repeated  stresses,  but  merely  as  an  empirical  state- 
ment which  agrees  with  the  limiting  values  determined  by  experi- 
ment, and  which  will  give  approximately  intermediate  values. 

In  designing  a  bar  which  is  to  be  subject  to  an  enormous  num- 
ber  of  repetitions   of  stress,  ranging   from   P  to  P,   the  ratio 

P1  S' 

__  is  the  same  as  —  ,    and  formula  (15)  gives  the  unit-stress  J> 

Jr  o 

which  will  cause  rupture.  To  be  sure  of  safety  a  factor  of  security 
must  be  applied  ;  if  a  be  this  factor,  (15)  becomes  the  formula  for 
working  strength,  or, 

(I5y         5    _<_(,    +  1=/P  +  "+f-2ef'\ 

*•         iV  2C        7^  2^  P*) 

from  which  the  proper  unit-stress  may  be  computed.  The  factor 
of  security  a  is  here  usually  taken  the  same  as  the  factor  of  safety 
for  a  steady  load  where  there  is  no  range  of  stress. 


ART.    82.     WORKING  STRENGTH  FOR  REPEATED  STRESSES.  147 

For  example,  consider  a  kind  of  wrought  iron  for  which 
u  =  55  ooo,  e  =  25  ooo,  and  f  =  12  500.  Then  with  a  factor 
of  4,  formula  (15)  becomes, 


Here  P  is  the  minimum  and  Pthe  maximum  stress  to  which  the 
piece  is  to  be  subjected,  the  first  due  perhaps  to  dead  load  and 
the  second  to  combined  dead  and  live  load.  The  following  are 
values  of  Sw  for  certain  ratios  of  P  to  P. 

For       ~  =  i,  Sw  =  13  750, 

For      -p  =  rt,  Sw=n46o, 

P 
For      --  =  y2,  Sw  =    9  400, 

For      ~=y4,  Sw=    7715, 

For      —3=0,  Sw  =    6  250, 

For  =   -  ^,  5W  =    4  720, 


For      -=  =  —  ^3,  Sw  =    4  030, 

For      Pp=—  l>  S«,=--    3  125. 

For  the  first  four  values  P  and  P  are  both  tensile  or  both  com- 
pressive,  and  in  the  last  three  values  P  is  the  reverse  of  P.  If  P 
be  tension  the  computed  values  of  5  are  to  be  used  at  once  in 
formula  (i)  for  finding  the  cross-sections,  but  if  Pbe  compression 
the  length  of  the  piece  should  be  taken  into  account  by  formula 
(10)  if  necessary. 

As  a  first  example,  let  it  be  required  to  find  the  proper  cross- 
section  of  a  wrought  iron  bar  which  is  to  be  subjected  to  a  re- 


148  APPENDIX    AND   TABLES.  VIII. 

peated  tension   ranging  from  30  ooo  to  90  ooo  pounds.     Here 
Pp-  =  X,  and, 


=  6250(1+  g  +-£;)=  8  250. 


Then  the  cross-section  of  the  bar  is  ?-  --  —  1  1  square  inches 

8  250 

For  a  second  example,  let  it  be  required  to  find  the  cross-section 
of  a  wrought  iron  bar  which  is  to  be  subjected  to  repeated  stress 
ranging  from  30  ooo  pounds  compression  to  90  ooo  tension. 

P 

Here  —  —  _   %,  and  from  the  formula  Sw  =  4  720.     Then 

the  cross-section  should  be  ?  --  —  19  square  inches. 
4720 

As  a  third  example,  the  cross-section  of  a  wrought  iron  bar  is 
required  when  the  stress  ranges  from  30  ooo  pounds  compression 
to  90  ooo  compression.  Here  as  before  S^  =  8  250.  This  value 
is  now  to  be  placed  for  Sc  in  Gordon's  formula,  and  the  cross- 
section  may  then  be  found  as  in  Art.  61,  for  any  given  length. 

The  quantity  /  which  is  the  unit-stress  required  to  produce 
rupture  after  an  enormous  number  of  repetitions  in  alternating 
stress  of  equal  amplitudes,  was  called  the  '  vibration  strength  '  by 
Wohler.  Its  value  for  wrought  iron  is  about  one-half  and  for 
steel  a  little  greater  than  one-half  the  elastic  limit.  For  cast  iron 
u  and  e  are  greater  in  tension  than  in  compression  and  this  should 
be  borne  in  mind  when  using  formula  (15). 

Prob.  132.  A  steel  bar  one  inch  in  diameter  is  subject  to  re- 
peated stress  ranging  between  1  5  ooo  pounds  tension  and  40  ooo 
pounds  tension.  Will  it  break  after  an  enormous  number  of 
repetitions  ? 

Prob.  133.  Find  the  proper  cross-section  for  a  cast  iron  bar  12 
feet  long  when  subjected  to  repeated  tension  ranging  from  30  ooo 
to  90  ooo  pounds.  Also  its  cross-section  when  subjected  to  re- 
peated compression  ranging  between  the  same  limits, 


ART.  83.  THE  RESILIENCE  OF  MATERIALS.  149 

ART.  83.     THE  RESILIENCE  OF  MATERIALS. 

When  an  applied  stress  causes  a  deformation  or  strain  work  is 
done.  Thus  if  a  tensile  stress  P  be  applied  by  increments  to  a 
bar,  so  that  the  stress  gradually  increases  from  o  to  the  value  P, 
the  work  done  is  the  product  of  the  average  stress  by  the  total 
elongation  X.  This  product  is  termed  the  resilience  of  the  bar. 
If  the  stress  does  not  exceed  the  elastic  limit  of  the  material  the 
average  stress  is  %P,  and  the  work  or  resilience  is  ^j/3/.  If  the 
cross-section  of  the  bar  be  A  and  its  length  /,  the  unit-stress  is 

P  ) 

_  or  S,  and  the  unit-strain  is  -  or  s,  so  that  the  work  done  on 
A  I 

each  unit  of  length  of  the  bar  per  unit  of  cross-section  is 
From  formula  ( 
may  be  written, 


From  formula  (2)  the  value  of  s  is  — ,  and  accordingly  this  work 


If  6"  be  the  unit-stress  at  the   elastic   limit,  the   quantity  K  is 
called  the  modulus  of  resilience  of  the  material. 

Resilience  is  a  measure  of  the  capacity  of  a  material  to  with- 
stand shock,  for  if  a  shock  or  sudden  stress  be  produced  by  a  fall- 
ing body,  its  intensity  depends  upon  the  weight  and  the  height 
through  which  it  has  fallen,  that  is,  upon  its  kinetic  energy  or 
work.  The  higher  the  resilience  of  a  material  the  greater  is  its 
capacity  to  resist  shocks.  The  modulus  of  resilience  is  a  measure 
of  this  capacity  within  the  elastic  limit  only. 

The  following  are  values  of  the  modulus  of  resilience  as  com- 

puted from  (16)  by  the  use  of  the  average  constants  given  in  Art.  5. 

For  timber,  K  =    3.0  inch-pounds, 

For  cast  iron,  K  =     1.2  inch-pounds, 

For  wrought  iron,         K  =  12.5   inch-pounds, 

For  steel,  K  =  26.5  inch-pounds. 

The  ultimate  resilience  of  materials  cannot  be  expressed  by  a 


150  APPENDIX    AND   TABLES.  VIII. 

rational  formula,  because  the  law  of  increase  of  elongation  beyond 
the  elastic  limit  is  unknown.  In  Fig.  i  the  ultimate  resilience  is 
indicated  by  the  area  between  any  curve  and  the  axis  of  abscissas, 
since  that  area  has  the  same  value  as  the  total  work  performed  in 
producing  rupture.  For  timber  and  cast  iron  the  ratio  of  these 
areas  is  about  the  same  as  that  of  the  values  of  K,  but  for  wrought 
iron  and  steel  the  areas  are  nearly  equal. 

Prob.  134.  What  horse-power  engine  is  required  to  strain  125 
times  per  minute  a  bar  of  wrought  iron  2  inches  in  diameter  and 
1 8  feet  long,  from  o  up  to  one-half  its  elastic  limit? 


ART.  84.     TABLES  OF  CONSTANTS. 

The  following  tables  recapitulate  the  mean  values  of  the  con- 
stants of  the  strength  of  materials  which  have  been  given  in  the 
preceding  pages.  It  is  here  again  repeated  that  these  values  are 
subject  to  wide  variations  dependent  on  the  kind  and  quality  of 
the  material,  and  for  many  other  reasons.  Timber,  for  instance, 
varies  in  strength  according  to  the  climate  where  grown,  the  soil, 
the  age  of  the  tree,  the  season  of  the  year  when  cut,  the  method 
and  duration  of  the  process  of  seasoning,  the  part  of  the  tree  used, 
the  knots  and  wind  shakes,  the  form  and  size  of  the  test  specimen 
and  the  direction  of  its  fibers,  so  that  it  is  a  difficult  matter  to 
state  definite  numerical  values  concerning  its  elasticity  and 
strength.  The  quality  of  the  material  causes  a  yet  wider  variation, 
so  wide  in  fact  that  in  some  cases  testing  machines  alone  could 
scarcely  distinguish  between  wrought  iron  and  steel ;  for  while 
the  higher  grades  of  steel  have  much  greater  strength  than  the 
tables  give,  the  mild  structural  and  merchant  steels  may  have 
values  almost  as  low  as  the  average  constants  for  wrought  iron. 
In  general,  therefore,  the  following  values  should  not  be  used  in 
actual  cases  of  investigation  and  design  except  for  approximate 
computations. 


ART.  84. 


TABLES. 


Detailed  tables  giving  the  results  of  experiments  upon  numer- 
ous kinds  and  qualities  of  materials  may  be  found  in  the 
following  books. 

Wood's  Resistance  of  Materials;  New  York,  1880. 
Burr's  Elasticity  and  Strength  of  Materials  ;  New  York,  1883. 
Thurston's  Materials  of  Engineering ;  New  York,  1884. 
Trautwine's  Engineers'  Pocket  Book  ;  New  York,  1885. 
Lanza's  Applied  Mechanics;  New  York,  1885. 


TABLE  I. 


Material. 

Mean  Weight. 

Coefficient  of  Linear 
Expansion. 

Pounds 
per  cubic 
foot. 

Kilograms 
per  cubic 
meter. 

For  i°  Fan. 

For  i  °  Cent. 

Timber 

40 

600 

O.OOOOO2O 

0.0000036 

Brick 

125 

2   OOO 

0.0000050 

0.0000090 

Stone 

160 

2    560 

0.0000050 

0.0000090 

Cast  Iron 

450 

7  200 

0.0000062 

O.OOOOII2 

Wrought  Iron 

480 

7  700 

0.0000067 

O.OOOOI2I 

Steel 

49° 

7  800 

0.0000065 

0.0000117 

TABLE   II. 


Elastic  Limit. 

Coefficient  of  Elasticity. 

Material. 

Pounds 

Kilograms 

Pounds 

Kilograms 

per  square 

per  square 

per  square 

per  square 

inch. 

centimeter. 

inch. 

centimeter. 

Timber 

3  ooo 

2IO 

i   500  ooo 

105  ooo 

Cast  Iron 

6  ooo 

420 

15  ooo  ooo 

i  050  ooo 

Wrought  Iron 

25  ooo 

i   750 

25  ooo  ooo 

i  750  ooo 

Steel 

40  ooo 

2    800 

30  ooo  ooo 

2    IOO   OOO 

152 


APPENDIX    AND    TABLES. 

TABLE   III. 


VIII. 


Material. 

Ultimate  Tensile 
Strength. 

Ultimate  Comparative  Strength. 

Pounds 

Kilograms 

Pounds 

Kilograms 

per  square 
inch. 

per  square 
centimeter. 

per  square 
inch. 

per  square 
inch. 

Timber 

10   000 

700 

8  ooo 

560 

Brick 

200 

H 

2  500 

175 

Stone 

6  ooo 

420 

Cast  Iron 

20  000 

I  400 

90  ooo 

6  300 

Wrought  Iron 

55  ooo 

3  850 

55  ooo 

3  850 

Steel 

100  000 

7  ooo 

150  ooo 

10  500 

TABLE  IV. 


Ultimate  Shearing 
Strength. 

Modulus  of  Rupture. 

Material. 

Pounds 

Kilograms 

Pounds 

Kilograms 

per  square 

per  square 

per  square 

per  square 

inch. 

centimeter. 

inch. 

centimeter. 

Timber 

f        600   | 
\    3  ooo   j 

f         42    1 

1     210  ; 

9  ooo 

630 

Stone 

2   OOO 

140 

Cast  Iron 

20  ooo 

i  400 

35  ooo 

2   450 

Wrought  Iron 

50  ooo 

3  5°o 

55  ooo 

3  850 

Steel 

70  ooo 

4  900 

I  2O   OOO 

8  400 

This  book  is  DUE  on  the  last  date  stamped  below 


JUN  1  1 


1931 


, 


1  1  195 
JUL  i:  ^ 


MAR   2     1961 


MAK3 
JNIAR1 


*"~ 


Form  L-9-15m-7,'32 


000  940  746     1 


